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3 years ago

How do you think the mass of an object affects how much friction get exerted on the object by a rough surface?

Physics

1 answer:

Shkiper50 [21]3 years ago

7 0

Explanation:

Friction can't affect mass of a substance(considering for a substance whose mass doesn't change with time),rather it is the mass of an object which can affect friction variously. Let's take some example to understand the situation. So,higher the mass of the object,higher will be the frictional force.

Friction occurs because no surface is perfectly smooth. Rougher surfaces have more friction between them. Heavier objects also have more friction because they press together with greater force. Friction produces heat because it causes the molecules on rubbing surfaces to move faster and have more energy.

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If the number of crests that pass a point in a given time increases, what has increased?

zysi [14]

The frequency increasing makes the crests pass more quickly. Frequency is a count of how many times per second an event occurs. In waves, this event is the passing of an entire cycle. Once the cycle has passed, the wave repeats. The faster the wave repeats, the higher the frequency. For this reason, frequency has units of hertz, Hz. The unit of hertz is 1/s or "per second"

7 0

3 years ago

Read 2 more answers

Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of ma

N76 [4]

Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

$c = \frac{x}{t}$

Where:

c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

Hence, the distance is:

$x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km$

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.

I hope it helps you!

5 0

3 years ago

A jet can travel the 6000 km distance between washington,

Andrei [34K]

<span> d = r*t is the basic distance equation
d = 6000 km
t with the tail wind = 6 hr
r with the tail wind = speed of the plane + wind speed = s + w
t with the head wind = 7.5 hr
r with the head wind = speed of the plane - wind speed = s-w
(s+w)*6 = 6000
(s-w)*7.5 = 6000
s + w = 1000
s - w = 800
</span><span> 2s = 1800
s = 900 km/h
s + w = 1000
w = 100
Check the anwer by calculating the return trip.
(900-100) * 7.5 = 800 * 7.5
800 * 7.5 = 6000 km
Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.</span>

6 0

3 years ago

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.