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olga_2 [115]
2 years ago
9

A cheetah starts from rest and accelerated at 8.7 m/s^2 for 3s. How far did the cheetah go in that time

Physics
1 answer:
nika2105 [10]2 years ago
4 0

Answer:

cheetah goes 52.2 m in that time.

Explanation:

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2 years ago
What layer of the atmosphere contains all of the weather and thus the most water vapor?
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The troposphere is the lowermost layer of the Earth's atmosphere. Most of the weather phenomena, systems, convection, turbulence and clouds occur in this layer, although some may extend into the lower portion of the stratosphere.

5 0
3 years ago
A television weighs 8.50 pounds. How many grams is this? (Hint: You need to
Ksenya-84 [330]

Answer:

3859 grams

Explanation:

Given: Weight of a television = 8.50 pounds

To find: Weight of a television in grams

Solution:

1 pound = 0.454 kg and 1 kg = 1000 g

So,

1 pound = 0.454 × 1000 = 454 grams

8.50 pounds = 8.50 × 454 = 3859 grams

Therefore,

Weight of television in grams = 3859 grams

7 0
3 years ago
Need help on this please
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The answer is D, the amount of energy stays the same.
3 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
2 years ago
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