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3 years ago

A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds. What is the car’s acceleration?

2 answers:

6 0

it should be A. 2 m/s² because the equation for acceleration is ( final− initial) / Time so you must subtract 10 from 40 and you'll get 30 then you divide by 15 to get your answer 2.

s344n2d4d5 [400]3 years ago

6 0

Answer:

The acceleration of the car is 2 m/s².

A is correct.

Explanation:

Given that,

Initial velocity =10 m/s

Final velocity =40 m/s

Time = 15 sec

The acceleration is the rate of change of velocity.

The acceleration of the car is

Using equation of motion

$v = u+at$

Here, v = final velocity

u = initial velocity

t = time

$40=10+a\times15$

$a = \dfrac{30}{15}$

$a = 2\ m/s^2$

Hence, The acceleration of the car is 2 m/s².

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10. If one body is positively charged and another body is negatively charged, free electrons tend to

marta [7]

Answer: B.move from the negatively charged body to the positively charged body.

Explanation: If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body. This happens because they have different charges and therefore electrons are more attracted to the positive or its opposite charge.

7 0

2 years ago

A 0.70-kg basketball dropped on a hardwood floor rises back up to 65% of its original height. (a) If the basketball is dropped f

dem82 [27]

Answer:

Part a)

$Loss = 3.6 J$

Part b)

$Loss = 0.99 J$

Part C)

This is loss in terms of thermal energy due to collision with the floor

Explanation:

Part a)

Since we know that the ball rises up by 65% of initial height

so after first bounce it will lose 35% of its initial energy

so we will have

$U = mgH$

Energy Loss = 0.35 mgH[/tex]

$Loss = 0.35(0.70)(9.81)(1.5)$

$Loss = 3.6 J$

Part b)

Energy of the ball after first bounce

$U_1 = 0.65 mgH$

energy of ball after 2nd Bounce

$U_2 = 0.65(0.65 mgH)$

energy of the ball after 3rd bounce

$U_3 = (0.65)(0.65^2)mgH$

$U_3 = 0.65^3(0.70)(9.81)(1.5)$

$U_3 = 2.83 J$

Now we will have energy loss in fourth bounce given as

$Loss = 0.35 U_3$

$Loss = 0.35(2.83)$

$Loss = 0.99 J$

Part C)

This is loss in terms of thermal energy due to collision with the floor

7 0

3 years ago

Subscribe to my channel

Troyanec [42]

Sure why not I'll sub your channel

3 0

3 years ago

write any two factors that affect the speed of sound in gaseous medium sound waves can't propagate in vaccum why

disa [49]

Explanation:

effect of density in speed of sound.
effect of direction of wind in speed of sound.

because there is no medium to travel sound in vacuum.

hope it helps.

7 0

3 years ago

A water pump that consumes 2 kW of electric power when operating is claimed to take in water from a lake and pump it to a pool w

Ket [755]

Answer:

The claim is not reasonable

Explanation:

From the question we are told that

The power consumed is $P = 2kW = 2*10^3 W$

The height of the pool's free surface is $h = 30 \ m$

The rate is $Q = 53\ L /s$

Given that the rate is 53 liters in 1 second , let say the 1 liter is equivalent to 1 kg so 50 liter = 50 kg

So the workdone by the pump is mathematically represented as

$W = m * g * h$

=> $W = 50 * 9.8 * 30$

=> $W =14700 \ J$

So what this value is telling us is that 14700 J of work is done in one second

and generally power is workdone per time thus the power consumed to move water at the given rate is 14.7 kW which is far greater than the power stated in the question so the claim is not reasonable

3 0

3 years ago