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3 years ago

How fast is the car traveling if it has driven a total of 200 miles in 5.5 hours?

1 answer:

6 0

Average speed = (distance covered) / (time to cover the distance)

   = (200 miles) / (5.5 hours)

   = (200/5.5) mile/hour

   = 36.4 miles per hour (rounded)

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You push a refrigerator with a force of 100 N. If you move the refrigerator a distance of 5 m while you are pushing, how much wo

ddd [48]

The work done to push the refrigerator is 500 Nm.

Explanation:

Work done is the measure of force required to move any object from one point to another. So it is calculated as the product of force and displacement.

If the force increases the work done will increase and similarly, the increase in displacement increases the work done. So to push the refrigerator work should be done on the object and not by the object.

As the force is 100 N and the displacement is 5 m then, work done can be measured as

Work = Force × Displacement

Work = 100 × 5 = 500 Nm

So the work done to push the refrigerator is 500 Nm.

7 0

3 years ago

A boy runs 400m at an average speed of 4.0m/s he runs the first 200m in 40 s how long does he take to run the second 200m?

IRISSAK [1]

If he runs at the same speed he will cover next 200m in 40s
that is at the average of 4.0m

8 0

4 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0

3 years ago

The police department is excited to have some new motorcycle units. One officer said that these motorcycles can go from 0 miles

vaieri [72.5K]

The officer is describing the motorcycle's acceleration

That is, he's describing how the motorcycle's velocity changes when compared to the time. That's the formula for calculating the average acceleration of any body.

3 0

3 years ago

A student measures the speed of sound by echo destiny classes hands and then measures the time to hear the echo his distance to

777dan777 [17]

Explanation:

∆x=300 m×2

∆t=1.5 s

v=∆x/∆t → v=2×300/1.5 = 400 m/s

6 0

3 years ago