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Delvig [45]
3 years ago
11

A cheetah runs with a positive acceleration. What does this tell you about the velocity time graph for the cheetah?

Physics
1 answer:
Katarina [22]3 years ago
3 0
B is the correct answer
You might be interested in
Help ASAP
EastWind [94]

Answer:

Elastic potential energy, E=2.35\times 10^{-8}\ J

Explanation:

Charge, q=9.4\times 10^{-10}\ C

Potential, V = 50 V

It is required to find the electric potential energy in a capacitor stored in it. The formula of the electric potential energy in a capacitor is given by :

E=\dfrac{1}{2}qV\\\\E=\dfrac{1}{2}\times 9.4\times 10^{-10}\times 50\\\\E=2.35\times 10^{-8}\ J

So, the electric potential energy stored in the capacitor is 2.35\times 10^{-8}\ J

8 0
3 years ago
A 5.0-kg centrifuge takes 95 s to spin up from rest to its final angular speed with constant angular acceleration. A point locat
stellarik [79]

Answer:

(a) 17.37 rad/s^2

(b) 12479

Explanation:

t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0

w = v / r = 99 / 0.06 = 1650 rad/s

(a) Use first equation of motion for rotational motion

w = w0 + α t

1650 = 0 + α x 95

α = 17.37 rad/s^2

(b) Let θ be the angular displacement

Use third equation of motion for rotational motion

w^2 = w0^2 + 2 α θ

1650^2 = 0 + 2 x 17.37 x θ

θ = 78367.87 rad

number of revolutions, n = θ / 2 π

n = 78367.87 / ( 2 x 3.14)

n = 12478.9 ≈ 12479

4 0
3 years ago
Who can help me?? physic question​
IgorC [24]

Assuming acceleration due to gravity of the moon is constant and there’s no initial velocity in the mans jump we can use one of the kinematic equations. x(final)=x(initial)+(1/2)gt^2. Plug in known values. 0=10-(1.62/2)t^2. The value 1.62 is acceleration of gravity on the moon. Now simply solve for t. t=3.513

5 0
3 years ago
Need help!!! A charge of 8.5 × 10–6 C is in an electric field that has a strength of 3.2 × 105 N/C. What is the electric force a
ddd [48]
B is the right answer. Multiply numbers you get the answer
3 0
3 years ago
Read 2 more answers
if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and
Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62

\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99

||v|| = 85

θ = 335°

\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03

\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92

Resultant

R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15

\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0
3 years ago
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