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Citrus2011 [14]
2 years ago
8

An ice skater accelerates backward for 5.0 second to final speed of 12m/s. If the acceleration backward was at a a rate of 1.5m/

s2 what was the skater's initial speed
Physics
1 answer:
Lisa [10]2 years ago
3 0

Answer:

Initial velocity, U = 4.5m/s

Explanation:

Given the following data;

Final velocity, v = 12m/s

Time, t = 5 seconds

Acceleration, a = 1.5m/s²

To find the initial velocity, we would use the first equation of motion.

V = U + at

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting into the equation, we have;

12 = U + 1.5*5

12 = U + 7.5

U = 12 - 7.5

Initial velocity, U = 4.5m/s

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When a balloon is deflating, why does air leave the balloon?
Gwar [14]

Answer:

<em>When a balloon deflates air moves out of the balloon </em><em>because the pressure inside the balloon is higher than the pressure outside the balloon.</em>

Explanation:

An inflated balloon has a high pressure region on its inside. Gases always move from a region of high pressure to a region of low pressure. When a balloon is inflated its membrane stretches making it even more porous.

The gas molecules inside the balloon easily diffuse out through this membrane. The diffusion rate may differ depending on the type of gas filled inside the balloon and the material of the balloon. For example helium balloon deflates faster than common air balloon.

This is because helium is a light element and can escape easier than gases like nitrogen and oxygen through the porous membrane of the balloon.

6 0
2 years ago
You have two vectors, which are 2.59 m at 30.0° north of east and 4.18 m at 60.0° north of west. What is the magnitude in meters
andrew11 [14]

Answer:\ec{r}=0.153\hat{i}+4.914\hat{j}v

Explanation:

Given

Vector 1

\vec{a}=2.59\left ( cos30\hat{i}+sin30\hat{j}\right )

Vector 2

\vec{b}=4.18\left ( -cos60\hat{i}+sin60\hat{j}\right )

Resultant \vec{r}=\vec{a}+\vec{b}

\vec{r}=2.59\left ( cos30\hat{i}+sin30\hat{j}\right )+4.18\left ( -cos60\hat{i}+sin60\hat{j}\right )

\vec{r}=0.153\hat{i}+4.914\hat{j}

|r|=4.916

5 0
3 years ago
What type of front occurs when cold air and warm air are next to each other but are at a standstill
Korvikt [17]

Answer:

A stationary front.

(Hope this helped! c:)

5 0
3 years ago
A race car's velocity increases from +28 m/s to +36 m/s over a 2.0-s time interval. What is the car's
BigorU [14]

Answer:

4 m/s^{2}

Explanation:

Acceleration, a=\frac {v-u}{t}

Where v and u are the final and initial velocities of the race car respectively, t is the time taken for the race car to attain velocity of 36 m/s.

Substituting 36 m/s for v, 28 m/s for u and 2 s for t then

a=\frac {36 m/s-28 m/s}{2}=4 m/s^{2}

3 0
2 years ago
A 35.2-g sample of an alloy at 93.0°C is placed into 219 g of water at 22.0°C in an insulated coffee cup. Assume that no heat is
Korvikt [17]

Answer:

Explanation:

mass of the alloy = 35.2 g

temperature t of the alloy = 93° C

heat loss by alloy = mc ( t₁ - t₂) where t₁ = 93 and t₂ = 31.1° C

heat loss by alloy = 2178.88 c where c is the specific heat capacity of the alloy

heat gained by water = m cw ( t₂ - t₁ ) where t₁ of water = 22° C and t₂ of water = 31.1° C

heat gained by water = 8338.294

heat gained = heat loss

2171.84 c =  8338.294

c specific heat capacity of the alloy = 8338.294 / 2178.88 = 3.83 J/g.K

7 0
3 years ago
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