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Elis [28]
3 years ago
12

A cheetah starts from rest and accelerates after a gazelle at a rate of 6.5 meters per second2for 3.0 seconds. Calculate the che

etah’s speed at the end of these 3.0 seconds.
Physics
1 answer:
Pepsi [2]3 years ago
6 0

Answer:

the speed of the cheetah at the end of the 3 seconds is: 19.5 m/s

Explanation:

Let's use the equation that relates speed with acceleration:

vf = vi + a * t

where vf stands for final velocity, vi stands for initial velocity, a for acceleration, and t for the time acceleration is applied. Then, in our case we have:

vf = 0 + 6.5 (3)

vf = 19.5 m/s

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Ulleksa [173]

Answer:

B and B

Explanation:

Question 1:

The distance is a unit of length, such as miles, meters, kilometers

The correct answer for this question would be 8m

Question 2:

Speed has a unit of length per unit of time.

The correct answer for this would be meters per second

-Chetan K

3 0
2 years ago
Density is calculated by combining 2 units, therefore the unit we use to measure density is called a _____________________unit.
nalin [4]

Answer: Dependent Unit or System of Units

Explanation:

Density is calculated by dividing mass (Kg) by volume (L).

The unit of Density is Kg/L or one of their derivatives such as g/cm³.

8 0
2 years ago
The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i
spayn [35]

Answer:

<em>a) 3.56 x 10^22 N</em>

<em>b) 3.56 x 10^22 N</em>

<em></em>

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = \frac{GMm}{R^2}

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = \frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2} = <em>3.56 x 10^22 N</em>

<em></em>

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>

b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>

7 0
3 years ago
A girl drops a stone from the top a tower 45m tall. At the same time, a boy standing at the base of the tower, projects another
Advocard [28]

Answer:

(i) The stones meet at 1.8 second

(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

Explanation:

(i)

First we consider the stone dropped by the girl. We have data:

Vi = Initial Velocity of Stone = 0 m/s   (Since the stone was initially at rest)

t = Time Period

g = 10 m/s²

s₁ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₁ = Vi t + (0.5)gt²

s₁ = (0)(t) + (0.5)(10)t²

s₁ = 5t²   ----- equation (1)

Now, we consider the stone throne vertically upward by the boy. We have data:

Vi = Initial Velocity of Stone = 25 m/s

t = Time Period

g = - 10 m/s²   (negative sign due to upward motion)

s₂ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₂ = Vi t + (0.5)gt²

s₂ = (25)(t) + (0.5)(-10)t²

s₂ = 25t - 5t²   ----- equation (2)

At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).

s₁ + s₂ = 45

using values from equation (1) and equation (2)

5t² + 25t - 5t² = 45

25t = 45

t = 45/25

<u>t =  1.8 sec</u>

(ii)

using this value of of t in equation (2)

s₂ = (25)(1.8) - (5)(1.8)²

<u>s₂ = 28.8 m</u>

using this value of of t in equation (1)

s₁ = (5)(1.8)²

<u>s₁ = 16.2 m</u>

<u>Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.</u>

6 0
3 years ago
Name and describe the apparatus used by Cavendish to discover the universal gravitation constant
Evgen [1.6K]
Its like a suspended wood with a lead sphere attached to each of its ends
3 0
3 years ago
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