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4 years ago

15

Each of the head lights in sammys car is connected to. A 12 v battery. When she switches on the headlamps, a current of 2a passe

s through The bulb calculate the power of the bulb

2 answers:

goldfiish [28.3K]4 years ago

8 0

Power = voltage x current.

Each light dissipates 24 watts of heat and light.

IRINA_888 [86]4 years ago

6 0

Answer:

its 45 watts 12x3.75

Explanation:

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You are in a boat that is 1400 kg. The current is 1500N and is pushing you back. If you have an acceleration of 3 m/s^2 what is

aleksandrvk [35]

Answer: The force of the engines is 5700N

Explanation:

By the second Newton's law, we have that:

F = m*a

Force equals mass times acceleration.

In this case, we have:

m = 1400kg

a = 3 m/s^2

And the force will be equal to the force of the engines, f, plus the force of the current, -1500 N (because this is pushing you back, so it is in the opposite direction than f), then we have:

F = f - 1500N

Then we have the equation:

f - 1500N = 1400kg*3m/s^2 = 4200N

f - 1500N = 4200N

f = 4200N + 1500N = 5700N

The force of the engines is 5700N

5 0

3 years ago

A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is atta

fredd [130]

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

$mu=(M+m)v$ (1)

where

$m=11.9 g = 11.9\cdot 10^{-3}kg$ is the mass of the bullet

$u=261 m/s$ is the initial velocity of the bullet

$M$ is the mass of the block

$v$ is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

$\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2$ (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

$v=\frac{mu}{M+m}$

And substituting into eq(2), we can solve to find the mass of the block:

$(M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg$

4 0

3 years ago

The graph below shows the velocity of a car as it attempts to set a speed record. At what point is the car the fastest?

Maksim231197 [3]

Answer:

B). 3.4 s

Explanation:

As we can see the graph is given between velocity and time

so here we can see that the velocity is changing here with time and initially for some time it moves with constant speed

Then it's speed decreases to next few second and then speed increases to its maximum value

The time after which velocity comes to its maximum value will reach after t = 3 s

so out of the all given options most correct option will be

$t = 3.4 s$

6 0

3 years ago

Which location(s) would usually have SURFACE WINDS blowing in TOWARDS IT from other places? *

Tems11 [23]

Number 2 is the anser tropics

4 0

3 years ago

Read 2 more answers

Several large firecrackers are inserted into the holes of a bowling ball, and the 6.3 kg ball is then launched into the air with

Anni [7]

Answer

given,

mass of the ball = 6.3 kg

speed of the ball = 10.4 m/s

angle made with horizontal = 43°

m_a = 1.8 kg v_a = 2.2 m/s

m_b = 1.6 kg v_b = 1.8 m/s

mass of third particle = 6.3 - 1.8 - 1.6

= 2.9 kg

u cos θ = 10.4 x cos 43° = 7.61 m/s

by using conservation momentum along x-axis

6.3 x 7.61 = 1.8 × (-2.2) + 0 + 1.6 × V₃ₓ

V₃ₓ = 32.44 m/s (toward right)

by using conservation momentum along y-axis

0 = 0 + 1.6 x 1.8 + 1.6 × V₃y

V₃y = -1.8 m/s (indicate downward)

velocity of the third particle

$v = \sqrt{32.44^2 + (-1.8)^2}$

v = 32.49 m/s

$tan \theta = \dfrac{-1.8}{32.44}$

$\theta = tan^{-1}(\dfrac{-1.8}{32.44})$

θ = 3.176° (downward with horizontal)

4 0

3 years ago