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2 years ago

A net force of 15 N is applied to a cart with a mass of 2.1 kg. What is the acceleration of the cart?

Physics

1 answer:

Shkiper50 [21]2 years ago

5 0

Force = mass * acceleration
15 = 2.1 * a
a = 15/2.1 = 7.143 m/s^2

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Block A of 5 kg with a speed of 3 m/s collides with block B of 10 kg that is stationary. After the collision, block B travels wi

Lemur [1.5K]

Answer:

-1m/s

Explanation:

We can calculate the speed of block A after collision

According to collision theory:

MaVa+MbVb = MaVa+MbVb (after collision)

Substitute the given values

5(3)+10(0) = 5Va+10(2)

15+0 = 5Va + 20

5Va = 15-20

5Va = -5

Va = -5/5

Va = -1m/s

Hence the velocity of ball A after collision is -1m/s

Note that the velocity of block B is zero before collision since it is stationary

6 0

3 years ago

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.

Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m$

The maximum height above the ground that the ball reaches is 7.09683 m

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s$

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

$v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s$

The velocity just before it hits the ground is 11.8 m/s

6 0

3 years ago

A wire has a current density of 6.25 × 10 6 A / m 2 6.25×106 A/m2 . If the cross-sectional area of the wire is 1.79 mm 2 1.79 mm

joja [24]

Answer:17.44A

Explanation: Current density=I/Area

Area is given by 2.79mm^2=2.79×10^-6m^2

Current=I=current density ×Area=6.25×10^6 ×2.79×10^-6=17.44A

3 0

3 years ago

A circuit has a current of 2.4 A. The voltage is increased to 4 times its original value, while the resistance stays the same. H

9966 [12]

Answer:

Resistance will become 4 times the previous value

Explanation:

We have given current in the circuit i = 2.4 A

According to ohm's law current in the circuit is given by $I=\frac{V}{R}$

So $2.4=\frac{V}{R}$ ............eqn 1

Now voltage is increased to 4 times so new voltage = 4 V

And current in the circuit is same as 2.4 A

We have to fond the resistance so that after increasing voltage current will be same

So $2.4=\frac{4V}{R_{unknown}}$ ..........eqn 2

Dividing eqn 1 and 2

$1=\frac{V}{R}\times \frac{R_{unkown}}{4V}$

$R_{unknown}=4R$

So resistance will become 4 times the previous value

6 0

3 years ago

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How much force is required to accelerate a 22Kg mass at 6 m/s?

GuDViN [60]

Answer:

F = 132N

Explanation:

7 0

3 years ago

Read 2 more answers