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vesna_86 [32]
3 years ago
12

If I were to take a ball to the moon, where gravity is 1/6th that on the Earth, what will happen to its GPE if I hold it at the

same height both places?
A. Not enough info
B. 6 times as much
C. 1/6th as much
D. Disappear
Physics
1 answer:
kodGreya [7K]3 years ago
7 0

Answer:

C

Explanation:

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Select all the expressions below which represent a 20% discount off the price of an item that originally costs d dollars.
dmitriy555 [2]

A - 0.8d

multiply the item (d) times 80% (which shows a 20% discount since 100%-20%=80%). Convert 80% to a decimal = .8

therefore, 0.8d

7 0
3 years ago
A 0.60-kilogram softball initially at rest is hit with a bat. The ball is in contact with the bat for 0.20 second and leaves the
Furkat [3]

Answer:

75 Newtons.

Explanation:

From Newton's second law of motion,

F = m(v-u)/t................... Equation 1

Where F = force exerted by the ball on the bat, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time

Given: m = 0.6 kilogram, u = 0 meter per seconds (at rest), v = 25 meters per seconds, t = 0.2 seconds.

Substitute into equation 1

F = 0.6(25-0)/0.2

F = 3(25)

F = 75 Newton.

Hence the magnitude of the average force exerted by the ball on the bat = 75 Newtons.

8 0
4 years ago
Suggest and explain two ways to reduce heat loss through the doors of the house.
emmainna [20.7K]
Keep the doors closed and make sure there isnt lots of open space between the door and the frame
8 0
4 years ago
A boat moves through the water with two forces acting on it. One is a 2.05 ✕ 103 N forward push by a motor, and the other is a 1
labwork [276]

Answer:

(a)  a= 0.139 m/s²

(b)  d= 4.45 m

(c) vf= 1.1 m/s

Explanation:

a) We apply Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass (kg)

a : acceleration (m/s²)

Data

F₁= +2.05 * 10³ N : forward push by a motor

F₂= -1.87* 10³ N : resistive force due to the water.

m= 1300 kg

Calculation of  the acceleration of the boat

We replace data in the formula (1):

∑F = m*a

F₁+F₂= m*a

a=\frac{F_{1} +F_{2} }{m}

a= \frac{2.05*10^{3} -1.87*10^{3}}{1300}

a= 0.139 m/s²

b) Kinematics of the boat

Because the boat moves with uniformly accelerated movement we apply the following formulas:

d= v₀t+ (1/2)*a*t²   Formula (2)

vf= v₀+at     Formula (3)

Where:  

d:displacement in meters (m)    

t : time interval (s)

v₀: initial speed (m/s)

vf: final speed (m/s)

a: acceleration (m/s² )

Data

v₀ = 0

a= 0.139 m/s²

t = 8 s

Calculation of the distance traveled by the boat in 8 s

We replace data in the formula (2)

d= v₀t+ (1/2)*a*t²

d= 0+ (1/2)*(0.139)*(8)²

d= 4.45 m

c) Calculation of the  speed of the boat in 8 s

We replace data in the formula (3):

vf= v₀+at

vf= 0+( 0.139)*(8)

vf= 1.1 m/s

6 0
4 years ago
One person drives 60,000 miles in a car that averages 30 miles per gallon (mpg), while another person drives 60,000 miles in a c
Solnce55 [7]

Answer:

20000 pounds

Explanation:

Generally 1 gallon of fuel burnt gives of 20 pounds of CO₂

Difference in the gallons of fuel burnt

\dfrac{60000}{20}-\dfrac{60000}{30}=1000\ gal

The 20 mpg car burns 1000 gallons of fuel more than the 30 mpg car

The amount of CO₂ would be

1000\times 20=20000\ pounds

The fewer pounds that the 30 mpg car would release is 20000 pounds

4 0
3 years ago
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