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Sergio039 [100]

2 years ago

11

A car is traveling 50mph how long will it take the car to travel 300 miles

1 answer:

baherus [9]2 years ago

4 0

50 mph = 50 miles in one hour
300 divided by 50 = 6 hours

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Peter throws a snowball at his car parked in the driveway. The snowball disintegrates as it hits the car. By Newton’s third law,

tamaranim1 [39]

"he force exerted by the car is more than the force exerted b y the snowball" is the one among the following that can be said <span>about the magnitude of the forces exerted by the snowball and the car. The correct option among all the options that are given in the question is the first option or option "A". I hope it helps you.</span>

3 0

3 years ago

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A submarine has a "crush depth" (that is, the depth at which

sergey [27]

Answer:

P =40.69 atm

Explanation:

We need to find the approximate pressure at a depth of 400 m.

It can be calculated as follows :

P = Patm + ρgh

Put all the values,

$P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm$

So, the approximate pressure is equal to 40.69 atm.

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2 years ago

In countries where there is very hot and very cold weather, should the gaps between the lengths of railway track be wider or nar

kipiarov [429]

They should be bigger as if it is hot the tracks will expand as it is made of metal and then there is more room for it to expand or contract

5 0

3 years ago

A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute

levacccp [35]

Explanation:

The given data is as follows.

Mass, m = 75 g

Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

where, $m_{1}$ = mass of the projectile

$m_{2}$ = mass of block

v = velocity after the impact

Now, putting the given values into the above formula as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

$75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v$

= $\frac{45}{50.075}$

v = 0.898 m/s

Now, equation for energy is as follows.

E = $\frac{1}{2}mv^{2}$

= $\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}$

= 13500 J

Now, energy after the impact will be as follows.

E' = $\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}$

= 20.19 J

Therefore, energy lost will be calculated as follows.

$\Delta E$ = E E'

= (13500 - 20) J

= 13480 J

And, n = $\frac{\Delta E}{E}$

= $\frac{13480}{13500} \times 100$

= 99.85

= 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

7 0

3 years ago

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Which refers to the amount of heat required to change the temperature of 1 gram of a substance by 1°C and is related to the chem

timama [110]

Answer:

Specific heat

Explanation:

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2 years ago