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3 years ago

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HW 5.2.A 5.00-kg chunk of ice is sliding at 12.0 m/s on the floor of an ice-covered valley whenit collides with and sticks to an

other 5.00-kg chunk of ice that is initially at rest. Since the valleyis icy, there is no friction. After the collision, how high above the valley floor will the combinedchunks go? (g= 9.8 m/s2)

1 answer:

Allisa [31]3 years ago

3 0

The concept used to solve this problem is the conservation of momentum and the conservation of energy.

For conservation of the moment we have the definition:

$m_1v_1 = (m_1+m_2)v_f$

Where,

m = Mass

$v_1$ = Initial velocity for object 1

$v_f$ = Final velocity

Replacing the values we have to,

$m_1v_1 = (m_1+m_2)v_f$

$5*12=(5+5)v_f$

$v_f = 6m/s$

By conservation of energy we know that the potential energy is equal to the kinetic energy then

$mgh = \frac{1}{2} m(v_f^2-v_i^2)$

$gh = \frac{1}{2} v_f^2$

$h = \frac{1}{2} g*v_f^2$

$h = \frac{1}{2} (9.8)(6)^2$

$h = 1.837m$

Therefore after the collision the height when the combined chinks will go is 1.837m

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The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

3 years ago

Anyone know these questions?

salantis [7]

400m in 32sec: (400/32)>12.5meters per second>
(12.5)(60)(60)(1/1000)=45km per hour
Constant speed would mean that the two forces are equivalent

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3 years ago

A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats

Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

$\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}$

<u>Where:</u>

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

$\lambda = \frac{c}{f}$

<u>Where:</u>

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

$\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m$

Hence, the distance is:

$d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m$

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

I hope it helps you!

7 0

3 years ago

A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi

malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

1)

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude $mg$ , where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

2)

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

$F_{net}=ma$

where

$F_{net}$ is the net force

m = 75.0 kg is the mass of the bicyclist

$a=2.20 m/s^2$ is its acceleration

Solving, we find the net force:

$F_{net}=(75.0)(2.20)=165 N$

3)

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

$F_{net}=F-R$

where:

F is the forward force of push

R is the air drag

We know that:

$F_{net}=165 N$ is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

$F=F_{net}+R=165+60=225 N$

6 0

3 years ago

Find an expression for the square of the orbital period.

jenyasd209 [6]

Answer:

T²= 4π²R³/GM

Explanation:

First we know that

Fg= Fc

Because centripetal force must equal gravitational force

So

GMm/R² = Mv²/R

But velocity is 2πR/T

So by substitution we have

GMm/R²= M (2πR/T)/T

We have

T²= 4π²R³/GM as period

8 0

3 years ago