How can you take my step sis out the washer machine?

A marble statue has a mass of 6,200 grams and a volume of 2,296 cm3. What is the density of marble?

Elan Coil [88]

D= m/v
d= 6200/2296
density = about 2.7

6 0

4 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n

satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0

4 years ago

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! A IS NOT THE CORRECT ANSWER

patriot [66]

Answer: A

<u>Explanation:</u>

NOTES:

d = 650 meters

t = 10 seconds

**********************************

v = d/t

= 650 meters/10 seconds

= 65 meters/second

6 0

4 years ago

Read 2 more answers

On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik

Anon25 [30]

a) $4.62\cdot 10^{14} J$

b) 0.110 megatons

c) 8.46 bombs

Explanation:

a)

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

$\Delta K=K_f-K_i$

Which can be rewritten as:

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where:

$m=3.2\cdot 10^6 kg$ is the mass of the meteorite

$v=0$ is the final speed of the meteorite

$u=17 km/s = 17,000 m/s$ is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

$\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J$

So, the energy lost by the meteorite is $4.62\cdot 10^{14} J$

b)

The energy equivalent to 1 megaton of TNT is

$E_{TNT}=4.2\cdot 10^{15} J$

Here the energy lost by the meteorite is

$E=4.62\cdot 10^{14} J$

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

$\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110$

So, the energy lost by the meteorite corresponds to 0.110 megatons.

c)

The energy of one atomic bomb explosion in Hiroshima is equal to

$E'=13 kt$ (13 kilotons)

which corresponds to

$E'=0.013 Mt$ (0.013 megatons)

Here the energy of the meteorite is equal to

$E=0.110 Mt$ (0.110 megatons)

Therefore, we can find how many Hiroshima bombs are equivalent to teh meteorite impact by using the following rules of three:

$\frac{1 bomb}{0.013 Mt}=\frac{x bombs}{0.110 Mt}\\x=\frac{1\cdot 0.110}{0.013}=8.46$

So, 8.46 bombs.

5 0

3 years ago

A train has a constant speed of 10 m/s around a track with a diameter of 45 m what is the centripal acceleration?

Bumek [7]

Answer:

$a_{c}= 4.44\frac{m}{s}$

Explanation:

When an object goes on a circular movement, it has a centripetal acceleration that always points toward the center of the circle, it is the responsible of the change of direction in the movement of the object. and that centripetal acceleration is related with the speed in the next way:

$a_{c}=\frac{v^{2}}{r}$ , with v the speed, r the radius of the track that is half of the diameter (22.5 m)

$a_{c}=\frac{10^{2}}{22.5}$

$a_{c}= 4.44\frac{m}{s}$

5 0

4 years ago