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Musya8 [376]
3 years ago
13

A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/

s. what will be the momentum of this ball just as it reaches the ground
Physics
1 answer:
Anastasy [175]3 years ago
3 0

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

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Answer:

Explanation:

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6 0
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A horizontal 745 N merry-go-round of radius
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Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

Substituting the values

Moment of Inertia of solid cylinder I  

=>0.5 \times 76.02 \times (1.45)^2

=> 0.5 \times 76.02\times 2.1025

=> 79.91 kg.m^2

Step 3: Finding the Torque applied T

Torque applied T = F \times r

Substituting the values

T = 56.3  \times 1.45

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,\alpha  = \frac{Torque}{Inertia}

Substituting the values,

\alpha  = \frac{81.635}{79.91}

\alpha = 1.021 rad/s^2

 Step 4: Finding the Final angular velocity

Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

\omega = 1.021 \times  3.62

\omega = 3.69 rad/s

Now KE (100% rotational) after 3.62s is:

KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

6 0
3 years ago
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Answer: A

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A flat circular coil having a diameter of 25 cm is to produce a magnetic field at its center of magnitude, 1.0 mT. If the coil h
tresset_1 [31]

Answer:

The current pass through the coil is 6.25 A

Explanation:

Given that,

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Using the formula of magnetic field

B =\dfrac{\mu_{0}NI}{2\pi r}

I=\dfrac{B\times2\pi r}{\mu N}

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Put the value into the formula

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castortr0y [4]

Answer:

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(i) rapid freezing,

(ii) fracturing,

(iii) replication and

(iv) replica cleaning.

4 0
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