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2 years ago

a car has a mass of 200kg. It is on a hill 1000m high. How much gravitational potential energy does the car have?

Physics

1 answer:

wariber [46]2 years ago

7 0

Answer:

Explanation:

PE = m * g * h

m = 200 kg

h = 1000 m

g = 9.81

PE = 200 * 9.81 * 1000

PE = 1962000 Joules

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The equation for distance is d = st. If a car has a speed of 85 m/s and travels

Ludmilka [50]

Answer:

D) 10 200 m

Explanation:

Speed is known as the rate of change of distance.

$speed=\frac{distance}{time}$

distance = speed * time

= 85 ms⁻¹ * 120 s

= 10200 m

4 0

2 years ago

What type of wave is produced when you move one end of a horizontal spring up and down?

Volgvan

A. a transverse wave is produced

7 0

2 years ago

Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg.

luda_lava [24]

A )
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81 + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487
26.487 - 15.2035 = (Mi) · 2.7 · 9.81 · 0.819
11.2835 = (Mi) · 21.69
(Mi) = 11.2835 : 21.69 = 0.52

4 0

2 years ago

A parallel-plate capacitor is connected to a battery until it is fully charged. Then, the capacitor is disconnected from the bat

Helga [31]

Answer:

The potential between the plates will decrease.

Explanation:

An insulator is usually placed between the parallel plates and is also called a dielectric because it makes the amount of charge a capacitor can accommodate to increase at a particular potential difference.

Furthermore, the dielectric effect will make the electric field of the charged capacitor which is not connected to a source of supply to decrease.

Now, when the battery is removed, the charge Q remains constant and Capacity C will increase.

Formula for the potential difference is here;

V = Q/C

Since the numerator Q is constant and the denominator C increases, it means the potential difference V will decrease

4 0

2 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

ELEN [110]

With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$

5 0

2 years ago