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romanna [79]
3 years ago
13

A stone is placed in a graduated cylinder to determine its volume. The following data are obtained. What is the volume of the st

one? The initial volume of water is 25.0 mL. The final volume of water is 29.5 mL
a) 4.50 cm 3
b) 4.50 m 3
c) 4.50 dm 3
d) 4.50 cm
Physics
2 answers:
LUCKY_DIMON [66]3 years ago
3 0
To solve this, you subtract the initial volume from the final volume. So, 29.5 mL minus 25.0 mL is 4.5 mL. Technically, at this point you need to say 4.50 mL, because of sig figs. The volume of the stone will not be in mL, because the stone is not a liquid or a gas, so you must convert it to cm3. So, your answer is (a), 4.50 cm3.
Hope this helps!
Ostrovityanka [42]3 years ago
3 0
A. is the Answer Cheers!
AlexFray555
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A dockworker applies a constant horizontal force of 73.0 N to a block of ice on a smooth horizontal floor. The frictional force
vivado [14]

Answer:

a) 57.0 kg b) 24.2 m

Explanation:

a) According Newton's second law, the applied force is equal to the product of the mass times the acceleration.

As the force is constant, the acceleration is constant too.

In this case, as we have as givens the distance and the time, and also we know that the block is starting form rest, we can get the acceleration as follows:

d = 1/2 * a * t² ⇒ a = 2d / t² ⇒ a= 2* 13.0 m / (4.5)² s² = 1.28 m/s²

Replacing in the Newton's 2nd Law equation:

F = m*a ⇒ m = F/a = 73.0 N / 1.28 m/s = 57.0 Kg

b) At t=4.5 s, applying the definition of acceleration, we can get the value of the velocity at that time, as follows:

v= a* t = 1.28 m/s * 4.5 s = 5.76 m/s

If the worker stops pushing at the end of the 4. 5 s, this means (neglecting friction) that from that time omwards, no net force acts on the block, so it continues moving at constant speed.

In order to get the distance moved in the next 4.20 sec, as it is moving at constant speed, we neeed just to apply the definition of velocity:

v= Δx / Δt  ⇒ Δx = v* Δt = 5.76 m/s * 4.2 m = 24.2 m

So, the total distance traveled during all the time (9.1 s) is just the sum of the 13.0 m advanced during the time when there was a constant force applied, and the last 24.2 m at constant speed, as follows:

d = 13.0 + 24.2 = 37.2 m

3 0
3 years ago
A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

5 0
3 years ago
When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racq
Lilit [14]

Answer:60 gm

Explanation:

Given

initial velocity of ball u=0

Force exerted by racquet F=540 N

time period of force t=5\ ms

final velocity of ball v=45\ m/s

Racquet imparts an impulse to the ball which is given by

J=F\Delta t=\Delta P

J=540\times \Delta t=m(45-0)

m=\frac{540\times 5\times 10^{-3}}{45}

m=60\ gm

8 0
4 years ago
Express 10^11 miners in examiners.<br> Answer in units of Emine ...?
balandron [24]
Since exa = 10^18, 10^11 miners are 0.0000001 Emine = 10^-7 Emine.
4 0
3 years ago
If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it take
kifflom [539]

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

8 0
3 years ago
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