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Novosadov [1.4K]

3 years ago

6

Please help me asdrtyuio

1 answer:

marta [7]3 years ago

4 0

Answer:

9. The Sun's Gravity

10. The core is the densest layer

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You push a coin across a table. The coin stops. How does this motion relate to balanced and unbalanced forces?

Snezhnost [94]

If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.

7 0

3 years ago

A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below t

maria [59]

Answer:

Explanation:

Suppose v is the initial velocity and $\theta$ is the angle of inclination

distance traveled in vertical direction in t=1 s

When gravity is present

$y=vt+\frac{1}{2}at^2$

where $y=vertical\ distance$

$a=acceleration$

$t=time$

$v=initial\ velocity$

here initial velocity is v\sin \theta [/tex] so

$y=v\sin \theta \times 1-\frac{1}{2}gt^2$

$y=v\sin \theta -0.5g$

In absence of gravity

$y_2=v\sin \theta \times t$

$y_2=v\sin \theta \times 1$

$\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m$

4 0

3 years ago

A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch

ANTONII [103]

Answer:

A. A = 0.913 m

B. amax = 132.24m/s^2

C. Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

$v_{max}=A\omega$ (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

$\omega=\sqrt{\frac{k}{m}}$ (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

$\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}$

Next, you solve the equation (1) for A and replace the values of vmax and w:

$A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m$

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

$a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}$

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

$F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N$

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0

3 years ago

Work is done on an object only if the force and displacement are __________?

miskamm [114]

It’s either movement or work or that’s what a quizzie said

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3 years ago

A visual poster can be used to helping the child?

creativ13 [48]

The correct answer is D. learn to do a task independently.

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3 years ago