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3 years ago

Answers for 13 are

Physics

1 answer:

Veseljchak [2.6K]3 years ago

5 0

A. Andesite

Members of the diorite family have an intermediate composition that is neither felsic nor mafic but has characteristics of both. ... Diorite, a coarse-grained rock, has less quartz than granite and less plagioclase feldspar than gabbro. Andesite is a fine-grained member of the diorite family.

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10 N pushes a 10 kg crate to the right. Determine the acceleration of the crate.

oksano4ka [1.4K]

The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1

8 0

3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

2 years ago

Read 2 more answers

A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th

vladimir1956 [14]

Answer:

Force, $F=2.27\times 10^4\ N$

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

$F.d=\dfrac{1}{2}m(v^2-u^2)$

Finally, it stops, v = 0

$F.d=-\dfrac{1}{2}m(u^2)$

$F=\dfrac{-mu^2}{2d}$

$F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}$

F = -22713.92 N

$F=2.27\times 10^4\ N$

So, the magnitude of the force that stops the bullet is $2.27\times 10^4\ N$

7 0

2 years ago

I don’t understand I need help asap

hammer [34]

For the first question, you got them right, for the two you left blank, initial(beginning) velocity: 2 m/s the final velocity is: 12 m/s

3 0

3 years ago

n a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic

Annette [7]

Answer:

Explanation:

a ) starting from rest , so u = o and initial kinetic energy = 0 .

Let mass of the skier = m

Kinetic energy gained = potential energy lost

= mgh = mg l sinθ

= m x 9.8 x 70 x sin 30

= 343 m

Total kinetic energy at the base = 343 m + 0 = 343 m .

b )

In this case initial kinetic energy = 1/2 m v²

= .5 x m x 2.5²

= 3.125 m

Total kinetic energy at the base

= 3.125 m + 343 m

= 346.125 m

c ) It is not surprising as energy gained due to gravitational force by the earth is enormous . So component of energy gained due to gravitational force far exceeds the initial kinetic energy . Still in a competitive event , the fractional initial kinetic energy may be the deciding factor .

7 0

3 years ago