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ValentinkaMS [17]
2 years ago
13

As planets move toward the sun, they tend to: A. move at constant speed B. speed up C. burn up D. slow down

Physics
1 answer:
dolphi86 [110]2 years ago
8 0

Answer:

B. speed up

Explanation:

The acceleration of an object due to a body's gravity is:

g = GM / r²

where G is the universal constant of gravitation,

M is the mass of the body,

and r is the distance from the body.

As a planet approaches the sun, r decreases.  As r decreases, g increases.

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a shopper pushes a cart 40.0m south down one aisle and then turns 90.0 degrees and moves 15.0m. He then makes another 90.0 degre
valentinak56 [21]

Answer:

Explanation:

The displacement is the distnce of the shopper from the starting point.

Sum of movement along the vertical = 40-20 = 20m

Movement along the horizontal (x direction) = 15.0m

Displacement will be gotten using the pythagoras theorem.

d = √20²+ 15²

d = √400+225

d = √625

d = 25.0m

Hence the shoppers total displacement is 25.0m

8 0
2 years ago
An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t
marissa [1.9K]
The energy transferred to the spring is given by:
U= \frac{1}{2}kx^2
where 
k is the spring constant
x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
k=52.9 N/cm = 5290 N/m
x=40.0 cm=0.4 m

so now we can use these data inside the equation ,to find the energy transferred to the spring:
U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J
4 0
2 years ago
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
A blue puck has a velocity of 3i –4j m/s. Its mass is 20 kg. What is its momentum?
damaskus [11]
P = m * v
v = {3i - 4j} = square root (3^2 + 4^2) = 5
P = 20 * 5
P = 100 kg m/s
6 0
2 years ago
Read 2 more answers
A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus
dexar [7]

Answer:

a=5m/s^2

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

v_{1}=150m/s

and the final velocity is :

v_{2}=0 m/s

so, the change in velocity is:

\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) =  150 m/s

and the change in time , according to the problem:

\Delta t=30s

So, the aceleration is:

a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2

6 0
2 years ago
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