A rock is thrown off a cliff with a speed of 5 m/s downward. How far will it fall after 7 seconds have elapsed?

If a metal has a mass of 78.2 g and a volume of 11.4 mL, what is its density?

hram777 [196]

Answer:

6.86g/mL

Explanation:

4 0

3 years ago

Consider the image above. Vi = the initial velocity and Vf = the final velocity. Is there acceleration? Explain your answer.

lutik1710 [3]

<span><span>Velocity is a vector, and the initial and final ones are in opposite directions.
There must have been acceleration in order to change the direction of motion.</span>

A) No. The initial and final velocities are the same.
This is all wrong, and not the correct choice.
It's "Yes", and the initial and final velocities are NOT the same.

B) Yes. The ball had to slow down in order to change direction.
This is poor, and not the correct choice.
The "Yes" is correct, but the explanation is bad.
Acceleration does NOT require any change in speed.

C) No. Acceleration is the change in velocity. The ball's velocity is constant.
This is all wrong, and not the correct choice.
It's "Yes", there IS acceleration, and the ball's velocity is NOT constant.

D) Yes. Even though the initial and final velocities are the same, there is a change in direction for the ball.
This choice is misleading too.
The "Yes" is correct ... there IS acceleration.
The change in direction is the reason.
The initial and final velocities are NOT the same. Only the speeds are.
</span>

3 0

3 years ago

Read 2 more answers

A 1-kg rock is suspended by a massless string from one end of a

maxonik [38]

Answer:

The weight of measuring stick is 9.8 N

Explanation:

given information:

the mass of the rock, $m_{r}$ = 1 kg

measuring stick, x =1 m

d = 0.25 m

to find the weight of measuring stick, we can use the following equation:

τ = Fd

τ = 0

$F_{r} d$ - $F_{s}d$ = 0

F_{r} = the force of the rock

F_{s} = the force of measuring stick

$F_{s} =F_{r}$

= m g

= 1 kg x 9.8 m/s

= 9.8 N

thus, the weight of measuring stick is 9.8 N

6 0

3 years ago

You wad up a piece of paper and throw it into the wastebasket. How far will

astraxan [27]

Answer:

Since the paper is wadded up tight, and if there's any

air resistance left we assume there isn't any, it might

just as well be a stone that's tossed. This is just a

stripped down projectile situation.

You said "an angle of 36 degrees", but you didn't say relative

to what. I'll assume that it's 36 degrees above horizontal, and

now I'll proceed to answer the question with the information that

I just gave myself.

-- The vertical component of the velocity is 1.4 sin(36)

= 0.823 m/s up.

-- The projectile rises for (0.823/9.8) second, runs out of gas,

and then falls for another (0.823/9.8) second to its original height.

So it's in the air for

2 (0.823/9.8) = 0.168 second

(not very long at all)

-- The horizontal component of the velocity is 1.4 cos(36)

= 1.133 m/s

and it doesn't change.

-- During the 0.168 second that it's in the air,

the wad travels horizontally

(0.168 s) x (1.133 m/s)

= 0.19 meter

(19 cm, ~ 7.5 inches)

If you find my mistake on this one, please please tell me.

As of now, it looks like with that velocity at that angle, your

paper wad only makes it 7.5 inches from your hand into the can.

Explanation:

6 0

3 years ago

Read 2 more answers

The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it

Keith_Richards [23]

Answer:

The no. of revolutions does the tub turn while it is in motion is = 52.51 revolutions

Explanation:

Given data

$\omega_1$ = 0

$\omega_2$ = 5 $\frac{rev}{sec}$

Time taken = 7 sec

(1). The angular acceleration is given by

$\alpha = \frac{d \omega}{dt}$

$\alpha = \frac{5}{7}$

$\alpha = 0.714 \frac{rev}{s^{2} }$

We know that from the equation of motion

$\omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_1$

$5^{2} = 0 + 2 (0.714) \theta_1$

$\theta_1 = 17.5 \ rev$ -------- (1)

(2). The angular acceleration is given by

$\alpha = \frac{d \omega}{dt}$

$\alpha = - \frac{5}{14}$

$\alpha =$ - 0.357 $\frac{rev}{s^{2} }$

We know that from the equation of motion

$\omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_2$

$0^{2} = 5^{2} + 2 (-0.357) \theta_2$

$\theta_2$ = 35.01 rev ------- (2)

Total no of revolution made by the machine is

$\theta = \theta_1 + \theta_2$

$\theta =$ 17.5 + 35.01

$\theta =$ 52.51 rev

Therefore the no. of revolutions does the tub turn while it is in motion is = 52.51 rev

8 0

3 years ago