A child in a boat throws a 5.90-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 25.0 kg and the mass of the boat is 38.4 kg. (Figure 1)
Calculate the velocity of the boat immediately after, assuming it was initially at rest.
Express your answer to three significant figures and include the appropriate units. Enter positive value if the direction of the velocity is in the direction of the velocity of the box and negative value if the direction of the velocity is in the direction opposite to the velocity of the box.

Answer:

44.7 seconds

Explanation:

D = 500 m

   500 = .25  t^2

    500/.25 = t^2

          2000 = t^2

                t = 44.7 seconds

In a transverse wave, the disturbance are carried forward and the particles are undergoing to and fro motion. Hence in transverse waves, there is no horizontal motion.

A wave can be described as a disturbance that travels through a medium from one location to the other location without transporting any matter.

There are two types of matter,

(i) Transverse wave

(ii) Longitudinal wave

In a transverse wave, the particles of the medium move in a direction perpendicular to the direction of propagation of the wave. That means, the particles undergo to and fro motion about their mean position and moves in the direction of the wave propagation.

Examples are:

Whereas longitudinal waves undergo back and forth motion with respect to the mean position.

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Answer:

The low energy X-ray calibration service is intended for thin-window plane-parallel chambers required for the dosimetry of superficial X-ray beams. Low-energy X-ray therapy is also known as superficial radiation therapy. This therapy uses beams of low-energy X-rays (radiation waves) to destroy skin cancer cells while not harming healthy tissue around or under the upper layers of skin.

The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.

In this question we must make use of Kirchhoff's laws to find the values of the missing currents in the circuit presented in the picture. There are two rules according to Kirchhoff's laws:

Based on the information given by the picture, we have the following system of linear equations that describes the entire circuit:

i₃ = i₁ + i₂    

- i₁ · R₁ + ε₁ - i₁ · r₁ - i₁ · R₅ + ε₂ - i₂ · (r₂ + R₂) = 0      

ε₂ - i₂ · (r₂ + R₂) - i₃ · r₄ - ε₄ - i₃ · r₃ + ε₃ - i₃ · R₃ = 0      

- i₁ - i₂ + i₃ = 0                                                      (1)

(R₁ + r₁ + R₅) · i₁ + (r₂ + R₂) · i₂ = ε₁ + ε₂                (2)

(r₂ + R₂) · i₂ + (r₄ + r₃ + R₃) · i₃ = ε₂ + ε₃ - ε₄        (3)

If we know that R₁ = 5 Ω, r₁ = 0.10 Ω, R₅ = 20 Ω, r₂ = 0.50 Ω, R₂ = 40 Ω, r₄ =  0.20 Ω, r₃ = 0.05 Ω, R₃ = 78 Ω, ε₁ = 22 V, ε₂ = 49 V, ε₃ = 10.5 V and ε₄ = 33 V, then the currents flowing in the circuit are:

- i₁ - i₂ + i₃ = 0

25.1 · i₁ + 40.5 · i₂ = 71

25.1 · i₂ + 78.25 · i₃ = 26.5

The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.

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The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]

a)

The center of mass of a three-particle system is expressed as

[tex]r_{cm}=[/tex] [tex]$$r_{c m}=\frac{\sum_{i=1}^{3} m_{i} r_{i}}{\sum_{i=1}^{3} m_{i}} \Rightarrow \frac{m_{1} r_{1}+m_{2} r_{2}+m_{3} r_{3}}{m_{1}+m_{2}+m_{3}}$$[/tex]

When the system is only on [tex]$\mathrm{x}$[/tex]-axis (i.e. [tex]$\mathrm{y}=0$[/tex] )

[tex]$$\begin{aligned}x_{c m} &=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\x_{c m} &=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\x_{c m} &=0.5909 \mathrm{~m} \\\text { Therefore } r_{c m}=(0.5909 \mathrm{~m}, 0)\end{aligned}$$[/tex]

b)

When the two particles are shifted

[tex]\begin{aligned}&x_{c m}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\&x_{c m}=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\&x_{c m}=0.5909 \mathrm{~m} \\&y_{c m}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}} \\&y_{c m}=\frac{(5.0 \times 1.0)+(2.0 \times 0)+(4.0 \times 0.5)}{5.0+2.0+4.0} \\&y_{c m}=0.6364 \mathrm{~m}\end{aligned}[/tex]

Therefore [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]

The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]

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This is a series of analysis of Angular velocity as relates to an asteroid's orbit. See the explanation below.

Asteroids are stony bodies that circle the Sun.

Although asteroids circle the Sun in the same way as planets do, they are far smaller.

Hence, from the information given:

A) The square of the period is proportional to the cube of the semimajor axis, according to Kepler's third law. As a result, the period of Y equals (E) the period of Z.

B) Angular momentum is preserved here, hence it is equivalent (E).

C) As eccentricity increases, so does the angular momentum. In this case, Y and Z have the same period, and both satellites cover the same proportion of the territory in the same length of time.

This indicates that a satellite on Z must cover a lesser area in a given period of time than a satellite on Y. The area swept is approximately 1/2 the radius times the tangential displacement.

Because both satellites have the same "radius" at point y, the satellite on Z must have a lower tangential velocity than the one on Y. As a result, Y has more angular momentum than (G) Z.

D) Using Kepler's third law, X's period is bigger than (G) Z's.

E) In a circular orbit, the angular velocity is constant. As a result, the angular velocity of Y at y equals (E) that at s.

F) Z's angular velocity at c is smaller than (L) at i.

G) Y's angular velocity at y is larger than (G) Z's at y.

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Answer:

Thermal conductivity formula

Explanation:

Answer: B

Explanation: i learned it last year

Pedometer is a small portable device that counts every step taken throughout the day. The correct option is (B).

A pedometer is a small portable device that counts the number of steps taken by an individual throughout the day. It is typically worn on the waist or carried in a pocket and uses a mechanism to detect body motion and count each step.

Pedometers are commonly used for tracking physical activity and encouraging individuals to achieve daily step goals as part of their fitness routines.

Therefore, Pedometer is a small portable device that counts every step taken throughout the day. Choose the option (B).

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(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) =  e^(-bx)

ln[-0.857 / cos(ω)] = -bx  

ln[-0.857 / cos(ω)] / b = - x  ---- (1)

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx  

ln[-0.57 / cos(2ω)] /2b = - x  ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) =  cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω  - cosω - 6 = 0

let cosω  = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω  = arc cos(-0.92)

ω  = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b  ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

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Answer:

The distance is

=

7

m

Explanation:

Apply the equation of motion

s

(

t

)

=

u

t

+

1

2

a

t

2

The initial velocity is

u

=

0

m

s

−

1

The acceleration is

a

=

2

m

s

−

2

Therefore, when

t

=

3

s

, we get

s

(

3

)

=

0

+

1

2

⋅

2

⋅

3

2

=

9

m

and when

t

=

4

s

s

(

4

)

=

0

+

1

2

⋅

2

⋅

4

2

=

16

m

Therefore,

The distance travelled in the fourth second is

d

=

s

(

4

)

−

s

(

3

)

=

16

−

9

=

7

m

2.89watts.

How much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m:

Formula: [tex]I\frac{P}{4\pi r^{2} }[/tex]

I=1.6x10-3 w/m2

r=12m

[tex]I\frac{P}{4\pi r^{2} }[/tex]

[tex]P=I4\pi r^{2}[/tex]

   [tex]=(1.6x10-3\frac{W}{m^{2} } ) (4\pi (12m)^{2} )[/tex]

The power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m  [tex]=2.89watts.[/tex]

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It is calculated that a) The angular velocity of the wheel is 272.13 rad/s,

b) On the edge of the grinding wheel, the linear speed is 47.62 m/s,

and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².

Calculation of angular velocity, linear speed & acceleration:

Provided that,

the diameter of the wheel = 0.35 m

So, the radius, r = 0.35/2 = 0.175 m

As 1 revolution = 2π rad

(a) the angular velocity, ω = 2600 rpm = [tex]\frac{2600 * 2\pi }{60}[/tex] rad/s

⇒ω = 272.13 rad/s

So, the angular velocity is 272.13 rad/s.

(b) The linear speed, v = r * ω

⇒v = 0.175 * 272.13

⇒v= 47.62 m/s

(c) The angular acceleration, [tex]a=\frac{v^{2} }{r}[/tex]

[tex]a = \frac{(47.62)^{2} }{0.175}[/tex]

⇒[tex]a[/tex] = 12958.08 m/s²

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Answer:

Approximately [tex]3.86\; {\rm N}[/tex] (given that the magnitude of this charge is [tex]-7.35\; {\rm \mu C}[/tex].)

Explanation:

If a charge of magnitude [tex]q[/tex] is placed in an electric field of magnitude [tex]E[/tex], the magnitude of the electrostatic force on that charge would be [tex]F = E\, q[/tex].

The magnitude of this charge is [tex]q = 7.35\; {\rm \mu C}[/tex]. Apply the unit conversion [tex]1\; {\rm \mu C} = 10^{-6}\; {\rm C}[/tex]:

[tex]\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}[/tex].

An electric field of magnitude [tex]E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}[/tex] would exert on this charge a force with a magnitude of:

[tex]\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}[/tex].

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

Given the following data:

First of all, we would calculate the electric field due to B₁ to B₃ and the electric field due to B₂ to B₃ respectively.

The electric field due to B₁ to B₃ is given by:

E₁ = kq₁/r₁²

E₁ = (9 × 10⁹ × 17 × 10⁻⁶)/(0.05² + 0.1²)

E₁ = 153 × 10³/0.0125

Electric field, E₁ = 12.24 × 10⁶ N/C.

The electric field due to B₂ to B₃ is given by:

E₂ = kq₂/r₂²

E₂ = (9 × 10⁹ × 5 × 10⁻⁶)/(0.1²)

E₂ = 45 × 10³/0.01

Electric field, E₂ = 4.5 × 10⁶ N/C.

Also, the magnitude of the angle formed is given by tan trigonometry:

Tanθ = 5/10

Tanθ = 0.5

θ = tan⁻¹(0.5)

θ = 26.56°.

Next, we would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₁:

E₁x = E₁cosθ - E₂

E₁x = 12.24 × 10⁶ × cos(26.56) - 4.5 × 10⁶

E₁x = 10.95 × 10⁶ - 4.5 × 10⁶

E₁x = 6.5 × 10⁶ N/C.

Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:

E₁y = -E₁sinθ

E₁y = -12.24 × 10⁶ × sin(26.56)

E₁y = -12.24 × 10⁶ × 0.4471

E₁y = -5.5 × 10⁶ N/C.

Also, the magnitude of the electric field is given by:

Exy = √(E₁x² + E₁y²)

Exy = √(6.5 × 10⁶)² + (-5.5 × 10⁶)²)

Exy = √4.225 × 10¹³ + 3.025 × 10¹³)

Exy = √(7.25 × 10¹³)

Exy = 8.5 × 10⁶ N/C.

We would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₂:

E₂x = -E₂sinθ

E₂x = -4.5 × 10⁶ × sin(90)

E₂x = -4.5 × 10⁶ N/C.

Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:

E₂y = E₂cosθ

E₂y = -4.5 × 10⁶ × cos(90)

E₂y = 0 N/C.

The magnitude of the net electric field at B₃ is given by:

Eₙ = √(E₁x + E₂x)² + E₁y²)

Eₙ = √(6.5 × 10⁶ + (-4.5 × 10⁶))² + (-5.5 × 10⁶)²)

Eₙ = √(2.5 × 10⁶)² + (3.025 × 10¹³)

Eₙ = √(6.25 × 10¹²) + (3.025 × 10¹³)

Eₙ = √(3.65 × 10¹³)

Eₙ = 6.04 × 10⁶ N/C.

For the direction, we have:

Tanθ = x/y

Tanθ = 6.5/-5.5

Tanθ = -1.1818

θ = tan⁻¹(-1.1818)

θ = 49.76°.

The magnitude of the force on B₃ due to this net electric field is given by:

F = B₃ × Eₙ

F = 26 × 10⁻⁶ × 6.04 × 10⁶

F = 157.04 Newton.

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Answer:

(a)If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision?

No. Because if you have initial momentum P⃗ ≠0 , if both of the objects were at rest after the collision the total momentum of the system would be P⃗ =0 , which violates conservation of momentum

(b)Is it possible for only one to be at rest after the collision?

Yes, that is perfectly possible. It characteristically, happens when both objects are of the same mass. When two objects of the same mass collide and Kinetic energy is conserved (Perfectly Elastic collision) then the two objects interchange velocities.

The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

Apply the following kinematic equation;

ωf² = ωi² - 2αθ

where;

ωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s

θ = 52 rev x 2π rad/rev = 326.7 rad

0 = ωi² - 2αθ

α = ωi²/2θ

α = ( 356.05²) / (2 x 326.7)

α = 194.02 rad/s²

Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

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Answer:

The term significant figures refers to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation . The number of significant figures in an expression indicates the confidence or precision with which an engineer or scientist states a quantity.

Answer:

each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first nonzero digit.

Explanation:

The force required to pull the two hemispheres is 46622.72N

( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]

The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.

Pressure difference = (940 - 12)

= 928 millibars.

(928 x 100)

= 92,800N/m^2.

Therefore, the required force to pull the two hemispheres is

(92800 x 0.5024)

= 46622.72N.

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Answer:

D.-4.798m/s

Explanation:

Greetings !

Given values

[tex]u= 4ms \\ a = 0.21ms {}^{2} \\ t = 3.8sec[/tex]

Solve for V of the given expression

Firstly, recall the velocity-time equation

[tex]v = u + at[/tex]

plug in known values to the equation

[tex]v = (4) + (0.21)(3.8)[/tex]

solve for final velocity

[tex]v = 4.792ms[/tex]

Hope it helps!

The ball orbit the Earth, when launched from the theoretical cannon of Newton, is option B. it is magnetically attracted.

Newton's cannonball was a hypothetical situation. Isaac Newton once proposed that gravity, which he believed to be a universal force, was the primary factor behind the planetary motion. In this experiment, Newton imagines projecting a stone or a cannonball onto the summit of a very tall mountain. The body should move away from Earth in the direction it was projected if there were no effects from gravity or air resistance.

Depending on the projectile's initial velocity and the gravitational force acting on it, the bullet will travel in a different direction. Low speeds result in a simple fallback to Earth. The Earth's surface causes the cannonball to deviate from its elliptical route.

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The wavelength of the wave is 1.16m and the velocity is 23.64m/s.

To find the answer, we have to know more about the Transverse waves.

                [tex]y(x,t)=0.048sin(5.40x-128t)[/tex]

               [tex]amplitude, a=0.048m\\wave vector, k=5.40\\angular frequency,w=128Hz[/tex]

                [tex]k=\frac{2\pi }{wavelength} \\\\wavelength=\frac{2\pi }{k} =\frac{2*3.14}{5.40} \\\\wavelength=1.16m[/tex]

                         [tex]v=frequency*wavelength\\2\pi f=w, thus,\\f=\frac{w}{2\pi } =\frac{128}{2*3.14}=20.38s^{-1}\\\\v=1.16*20.38=23.64m/s[/tex]

Thus, we can conclude that, the wavelength of the wave is 1.16m and the velocity is 23.64m/s

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A quantitative description of a system is important as it provides information about the carrying capacity of habitable worlds.

Quantitative description refers to the description of a system which is focused on the numerical value of the properties or components of the system.

For example, a quantitative description of the components of a given habitat will be focused on the number of the individual species in the habitat. It will also be focused on the numerical value of the non-living components of the system and how such values affect the living components of the system numerically.

This information will then be used by scientists to see how modifications of the various quantitative components of the habitat will help to improve the chances of survival of species found in the habitat. This leads to such concepts as the carrying capacity of a habitat which is the maximum number of species that the available resources is a habitat can easily sustain.

In conclusion, a quantitative description is important to in investigations about the habitable world.

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The final speed of the electron is 4.64 * 10^5 m/s.

Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.

E= F/q

F = Eq

F =  330 N/C * 1.6 * 10^-19 C

F = 5.28 * 10^-17 N

F = ma

a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg

a = 5.8 * 10^13 m/s^2

Using

v = u + at

u = 0 m/s because the electron was initially at rest

v = at

v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s

v = 4.64 * 10^5 m/s

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The length of the wire is 1.12m

What is Resistance ?

Resistance is opposition to the current flowing in the circuit.

Given ,

D= 0.6m

R= 0.6/2m

R= 28Ω

rho = 1Ωm

R= rho ×l/A

where R is resistance , l is length ,A is area and rho is resistivity

By applying value and Solving we get,

l= 1.12m

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Answer:

long-term

Explanation:

usually a short term goal is able to be accomplished in a week or two. the question gives a 3 month time frame for the person to build up to the end goal.

The vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.

As per conversation of momentum, mass of bullet × velocity= mass of pendulum × velocity

=> 0.029kg × 200 = 3.429 × velocity of pendulum

=> Velocity of pendulum= 5.8/3.429 = 1.7 m/s

1/2 × mass of pendulum × velocity²=1/2 × k × amplitude²

=> 10 = 13.44× amplitude²

=> Amplitude = √(10/13.44)

= 0.86m

Angle = tan inverse (amplitude/length of pendulum)

= tan inverse (0.86/2.5)

= 20°

Maximum displacement along vertical direction= L - L×cos20°

= 2.5 - 2.5×cos20°

= 0.15 m

Maximum displacement along horizontal direction= Lsin20°

= 2.5 × sin 20°

= 0.85m

Thus, we can conclude that the vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.

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Answer:

The red color in the sky at sunset (and sunrise) is due to an effect called Rayleigh scattering.

Explanation:

(1) The INCREASE in the sound level from the ambient work environment level (in dB) is 1 dB.

(2) The factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.

The INCREASE in the sound level from the ambient work environment level (in dB) is calculated as follows;

Increase in sound level = final sound level - original sound level

Increase in sound level = 86 dB - 85 dB = 1 dB

from the graph at 1.5 hours/day, sound level = 97 dB

Increase in sound intensity = final sound level - original sound level

Increase in sound intensity = 116 dB - 97 dB = 19 dB

Factor increase = 19/97 = 0.196 = 19.6%

Thus, the factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.

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Answer:

potential energy = (mass × height × acceleration due to gravity.)

750 = mass × 3 × 10

hence, mass = 750/30 = 25 kg.

Explanation:

first rule of parallel circuits : the voltage is the same in the whole circuit, no matter where we measure.

so, we have 1.5V at every bulb.

second rule of parallel circuits : the total current is the sum of the individual branch currents.

so, if the battery really allows it, then we have 3 times 0.4 A = 3×0.4 = 1.2 A as current in the circuit.