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muminat
3 years ago
12

Help with question b

Physics
1 answer:
givi [52]3 years ago
6 0

Answer:

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Explanation:

h

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A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continue
enot [183]

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

7 0
3 years ago
If you could travel at the speed of light, how long would it take to travel from our solar
Illusion [34]

Answer:

4.689 years

Explanation:

6 0
3 years ago
When we go up From the earth surface.what happens about the atmospheric pressure​
topjm [15]

Answer:

There are two reasons why air pressure decreases as altitude increases: density and depth of the atmosphere. Most gas molecules in the atmosphere are pulled close to Earth's surface by gravity, so gas particles are denser near the surface.

Explanation:

7 0
3 years ago
Describe in a couple sentences how the index of refraction is related to the speed of light?
lesantik [10]

Answer: n = c / v" "c" is the speed of light in a vacuum, "v" is the speed of light in that substance and "n" is the index of refraction. According to the formula, the index of refraction is the relation between the speed of light in a vacuum and the speed of light in a substance.

Explanation: the relation is the vacuum and the speed of light in a substance.

3 0
3 years ago
A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=
VARVARA [1.3K]

Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

U=\frac{1}{2}k (\Delta x)^2

where

k is the spring constant

\Delta x is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

\Delta x = d-0=d

and the amount of energy required is

U=\frac{1}{2}k d^2

In the second case, the spring is compressed from x=0 to x=-d, so

\Delta x = -d -0 = -d

and the amount of energy required is

U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2

so we see that the amount of energy required is the same.

7 0
3 years ago
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