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3 years ago

7

Otherwise, Newton's First Law of Motion says that an object will continue to stay at rest or at a _____ until a force is applied

to it.

2 answers:

andrezito [222]3 years ago

5 0

Your answer would be motion because u need force to stop motion or anything at that matter

padilas [110]3 years ago

3 0

However, the words "constant velocity" belong in the blank.

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Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st

DedPeter [7]

Answer:

E1 = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q) = +11nC

distance between thin glass rods = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = $\frac{Q}{2\pi e_{0}rL }$

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.01} - \frac{1}{0.03} )$ ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

= $\frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )$

= 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.02} - \frac{1}{0.02} )$

therefore E2 = 0

E1 = E3

hence E3 = 10.15*10^4 N/C

4 0

3 years ago

You are trying to climb a castle wall so, from the ground, you throw a hook with a rope attached to it at 24.1 m/s at an angle o

Serhud [2]

Answer:

The value is $h = 13.2 \ m$

Explanation:

From the question we are told that

The speed of the rope with hook is $u = 24 .1 \ m/s$

The angle is $\theta = 65.0^o$

The speed at which it hits top of the wall is $v = 16.3 m/s$

Generally from kinematic equation we have that

$v_y^2 = u_y ^2 + * 2 (-g)* h$

Here h is the height of the wall so

$[16.3 sin (65)]^2 = [24.1 sin (65)] ^2+ 2 (-9.8)* h$

=> $h = 13.2 \ m$

4 0

3 years ago

An artist wants to create a metal sculpture using a mold so that his artwork can be readily mass produced. He wants his sculptur

lukranit [14]

Answer:NO

Explanation:

No the mold should not be of the same size as that of sculpture because the material from which molds is made may shrink or expand depending upon its properties .

For example grey cast iron shrinks on cooling.

We need to make mold bigger in general so that if there is a need of finishing it can be done easily without altering the size of sculpture.

5 0

3 years ago

Choose all facts that increase the orbital velocity of a vessel around planet B. Bigger mass of planet B smaller mass of planet

telo118 [61]

Answer:

- Bigger mass of planet B

- orbiting closer to planet B

Explanation:

The orbital velocity of the vessel around the planet can be found by equalizing the force of gravity between the vessel and the planet and the centripetal force:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

m is the mass of the vessel

M is the mass of the planet

r is the distance between the vessel and the centre of the planet

v is the orbital velocity of the vessel

Re-arranging the formula, we find an expression for v:

$v=\sqrt{\frac{GM}{r}}$

We see that:

- the bigger the mass of the planet, M, the bigger the velocity

- the bigger the distance between the vessel and the planet, r, the smaller the velocity

So, the correct choices that increase the orbital velocity are:

- Bigger mass of planet B

- orbiting closer to planet B

6 0

3 years ago

A record turntable rotates through 5.0 rad in 2.8 s as it is accelerated uniformly from rest. What is the angular velocity at th

Usimov [2.4K]

Answer:

$\omega_f = 3.584\ rad/s$

Explanation:

given,

turntable rotate to, θ = 5 rad

time, t = 2.8 s

initial angular speed = 0 rad/s

final angular speed = ?

now, using equation of rotational motion

$\theta = \omega_i t + \dfrac{1}{2}\alpha t^2$

$5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2$

$\alpha= \dfrac{10}{2.8^2}$

α = 1.28 rad/s²

now, calculation of angular velocity

$\omega_f = \omega_i + \alpha t$

$\omega_f =0 +1.28\times 2.8$

$\omega_f = 3.584\ rad/s$

hence, the angular velocity at the end is equal to 3.584 rad/s

4 0

3 years ago