PHYSICS COLLEGE

A toy airplane is flying at a speed of 6 m/s with an acceleration of 0.3 m/s2How fast is it flying after 4 seconds?
A. 5.7 m/s
B. 2.6 m/s
C. 13.9 m/s
D. 7.2 m/s
SUBMIT

Answers

Answer 1

Answer:

Answer:

Explanation:

0.3=v-6/4

make v the subject

v=7.2ms^2

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When dots are further apart on a ticker-tape diagram, it indicates an object is moving
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Two violin strings are tuned to the same frequency 294 H. The tension in one string is then decreased by 2.0%. What will be the beat frequency heard when the two strings are played together?
As you take the stoppered part of the tube up the staircase you begin to see the water level drop around the 4th floor. As you continue up it does not continue up with you but stays at a constant level. What does that mean?a. The pressure in the tubing is equal to the barometric pressure.b. The tubing was unable to supply any more water to the tube for use.c. The pressure outside the tube is higher that the water pressure inside the tube.

What two statements are true about energy transformations

Answers

Answer:

First statement:

Energy can neither be created nor destroyed.

Second statement:

Energy can be converted from one form to another.

Explanation:

According to the law of conservation of energy:energy can neither be created nor destroyed but can be converted from one form to another

A centrifuge is a common laboratory instrument that separates components of differing densities in solution. This is accomplished by spinning a sample around in a circle with a large angular speed. Suppose that after a centrifuge in a medical laboratory is turned off, it continues to rotate with a constant angular deceleration for 10.0s before coming to rest.Part A

If its initial angular speed was 3890rpm , what is the magnitude of its angular deceleration? (|?| in revs/s^2 )

Part B

How many revolutions did the centrifuge complete after being turned off?

Answers

Answer:

$a_r=389\ rev.s^(-2)$

$n=58350 rev$

Explanation:

Given:

time of constant deceleration, $t=10\ s$

initial angular speed, $N_i=3890\ rpm\$

Using equation of motion:

$N_f=N_i+a_r.t$

$0=3890+a_r* 10$

$a_r=389\ rev.s^(-2)$

Using eq. of motion for no. of revolutions, we have:

$n=N_i.t+(1)/(2) a_r.t^2$

$n=3890* 10+0.5* 389* 100$

$n=58350\ rev$

What force causes oppositely charged particles to attract each other? A. Magnetic force B. Compression
C. Electrical Force D. Gravity

Answers

Answer:

It is electrical force

Explanation:

i got it wrong on A P E X with magnetic hope this helps!

Answer:

The electromagnetic force.

A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is the acceleration of the ride?4.0 m/s2
0.67 m/s2
0.075 m/s2
54 m/s2

Answers

4.0 m/s2

it's 9 squared divided by 6

the answer is B 4.0 m/s2

Which statement best describes how the first quatrain relates to the second quatrain? The first shows the beloved’s actions; the second describes how she imitates them. Both the first and the second show the actions of the speaker and the beloved. The first shows the speaker’s actions; the second shows the beloved’s opposition to them. The first shows the speaker’s sadness; the second shows the beloved’s anger.

Answers

Answer: the first shows the speakers actions; the second shows the beloveds opposition to them

Explanation:

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?

Answers

Answer:

a) $v=5.6725\,m.s^(-1)$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^(-2)$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^(-1)$

putting the values in eq. (1)

$v^2=0^2+2* 3.62 * 10$

$v=8.5088\,m.s^(-1)$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20* 8.5088$

$p=170.1764\,kg.m.s^(-1)$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)* v'=p$

$(20+10)* v'=170.1764$

$v'=5.6725\,m.s^(-1)$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$(1)/(2) (M+m).v'^2=(M+m).g.h$