All categories

4 years ago

A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?

1 answer:

5 0

The initial kinetic energy of the car is
$E_1 = \frac{1}{2}mv_1^2 = 1.9 \cdot 10^3 J$

Then, the velocity of the car is decreased by half: $v_2 = \frac{v_1}{2}$
so, the new kinetic energy is
$E_2 = \frac{1}{2}mv_2 ^2 = \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4}$
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
$E_2 = \frac{1.9 \cdot 10^3 J}{4}=475 J$

You might be interested in

Two racing boats set out from the same dock and speed away at the same constant speed of 104 km/h for half an hour (0.500 h), th

Zinaida [17]

Answer:

The blue boat traveled 6.1 km farther west than the green boat

The green boat traveled 10.7 km farther south than the blue boat

Explanation:

The equation for linear uniform speed movement is

X(t) = X0 + v * t

Since we have two coordinates (X, Y) we use

X(t) = X0 + vx * t

Y(t) = Y0 + vy * t

The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.

The blue boat moves with a direction 24° south of west, so it will have speeds:

vxb = 104 * cos(24) = 95 km/h

vyb = 104 * sin(24) = 42.3 km/h

And the green boat:

vxg = 104 * cos(37.7) = 82.3 km/h

vyg = 104 * sin(37.7) = 63.6 km/h

After half hour the boats will have arrived at positions

Xb = 95 * 0.5 = 47.5 km

Yb = 42.3 * 0.5 = 21.1 km

And

Xg = 82.3 * 0.5 = 41.4 km

Yg = 63.6 * 0.5 = 31.8 km

The difference in positions of the boats

47.5 - 41.4 = 6.1 km

31.8 - 21.1 = 10.7 km

5 0

3 years ago

What conversion factor is used to convert 20 cm to m

Anastaziya [24]

The conversion factor you use is 100 cm = 1 m.
You can divide 20 by 100 to get the answer.
20 cm/100 cm =.2 m
Hope this helped!

7 0

3 years ago

The data in the table above show how far

Taya2010 [7]

Answer:c

Explanation: because if you work it in a paper it should like lil wit is straight the numbers are going up by 16

6 0

3 years ago

A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of

Alexandra [31]

Answer:

The maximum no. of electrons- $2.25\times 10^{22}$

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

$I = \frac{Q}{t}$

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

$Q = 1.0\times 3600 = 3600\ C$

Maximum number of electrons, n is given by:

$n = \frac{Q}{e}$

where

e = charge on an electron = $1.6\times 10^{- 19}\ C$

Thus

$n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}$

3 0

3 years ago

A car is traveling with a velocity of 24.4 m/s. It accelerates at a constant rate of 3.2m/s2. If the acceleration lasts for 5.6

larisa86 [58]

Answer: 42.32 ms-1

Explanation:

v = u+ at

v = 24.4 + ( 3.2×5.6)

v = 42.32 ms-1

6 0

3 years ago