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Dafna11 [192]
3 years ago
11

A 53.4 kg ice skater is at rest when she throws a 0.135 kg snowball east at 15.8 m/s. What is her velocity afterwards?

Physics
1 answer:
il63 [147K]3 years ago
5 0

Answer:

Her velocity afterwards is 0.04 m/s to the west.

Explanation:

Using the conservation of momentum, we have that:

m1u1 + m2u2 = m1v1 + m2v2

Where m is the mass, u is the inicial velocity and v is the final velocity.

The inicial momentum is zero, because the ice skater and the snowball are at rest (u1 = 0 and u2 = 0)

The mass of the snowball is m1 = 0.135 kg and the final velocity is v1 = 15.8 m/s, so the momentum of the snowball is:

m1 * v1 = 0.135 * 15.8 = 2.133 kg*m/s

So from the conservation of momentum, we have that:

0 = 2.133 + m2v2

m2v2 = -2.133

The mass of the ice skater is m2 = 53.4 kg, then we have that:

53.4 * v2 = -2.133

v2 = -2.133 / 53.4 = -0.0399 m/s

Her velocity afterwards is 0.04 m/s to the west.

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An ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring
lozanna [386]

Answer:

0.6 m

Explanation:

When a spring is compressed it stores potential energy. This energy is:

Ep = 1/2 * k * x^2

Being x the distance it compressed/stretched.

When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.

The potential energy of the ice cube is:

Ep = m * g * h

This is vertical height and is related to the distance up the slope by:

sin(a) = h/d

h = sin(a) * d

Replacing:

Ep = m * g * sin(a) * d

Equating both potential energies:

1/2 * k * x^2 = m * g * sin(a) * d

d = (1/2 * k * x^2) / (m * g * sin(a))

d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m

8 0
3 years ago
How many 1140 nm long molecules would you have to line up end to end to stretch a distance of 158 miles?
dezoksy [38]

Answer:

221754385964.9123

Explanation:

Convert miles to nanometer

1 mile = 1.6 km

1 km = 1×10³×10³×10³×10³ nm

1 mile = 1.6×10¹² nm

So,

158 miles = 158×1.6×10¹² = 252.8×10¹² nm

Length of each molecule = 1140 nm

Number of molecules = Total length / Length of each molecule

\text{Number of molecules}=\frac{252.8\times 10^{12}}{1140}\\\Rightarrow \text{Number of molecules}=221754385964.9123

There are 221754385964.9123 number of molecules in a stretch of 158 miles

3 0
3 years ago
A virtual image is formed 17.0 cm from a concave mirror having a radius of curvature of 39.0 cm. (a) Find the position of the ob
Wittaler [7]

Explanation:

Image distance, v = -17 cm (-ve for virtual image)

Radius of curvature of concave mirror, R = 39 cm

Focal length, f = -19.5 cm (-ve for a concave mirror)

(a) Using mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{-19.5}-\dfrac{1}{(-17)}  

u = 132.6 cm    

So, the object is placed 132.6 cm in front of the mirror.

(b) Magnification of the  mirror, m=\dfrac{-v}{u}

m=\dfrac{-17}{132.6}

m = -0.128

Hence, this is the required solution.

7 0
3 years ago
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