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Neko [114]
3 years ago
7

If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.

Which statement best describes the force(s) acting on the football? A force must be applied to it when it slows to a position of rest. A force must be applied to it before it slows to a position of rest. No force is needed; an object in motion stays in motion. Many forces must act upon it to keep it in motion.
Physics
2 answers:
elena-14-01-66 [18.8K]3 years ago
6 0
Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>
Naily [24]3 years ago
6 0

Answer:

No force is needed; an object in motion stays in motion.

Explanation:

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A whistle you use to call your hunting dog has a frequency of 21 kHz, but your dog is ignoring it. You suspect the whistle may n
uranmaximum [27]

The Doppler effect is the right concept to solve this problem. The Doppler effect is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be described as,

f = (1-\frac{v_0}{v})f_0

Here,

f_0 = Frequency of the sound from the Whistle

f = Frequency of sound heard

v = Speed of the sound in the Air

Replacing we have that

1- \frac{v_0}{343} = \frac{20kHz}{21kHz}

\frac{v_0}{343} = 1-\frac{20}{21}

\frac{v_0}{343} = \frac{1}{21}

v_0 = \frac{1}{21}(343)

v_0 = 16.33m/s

Therefore the minimum speed to know if the whistle is working is 16.33m/s

3 0
3 years ago
A block of mass 3kg is placed on a rough surface. The coefficient of static friction between two surfaces is 0.2 then what is th
yarga [219]

Answer:

the minimun horizontal force is = 5,88 N

Explanation:

Using a free body diagram we can calculate this force, in the image attached and using Newton's law we have:

3 0
2 years ago
A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.796
Agata [3.3K]

Answer:

distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

Explanation:

Given that;

mass of block m = 0.200 kg

distance travelled d = 0.796 m

time t = 2.00 s

m₂ = 0.400 kg

If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?

Now, using the second equation of motion;

d = ut + (\frac{1}{2} × at²)

as the object started from rest, u=0

so, we substitute

0.796  = 0×2 + (\frac{1}{2} × a(2)²)

0.796  = 0 + (\frac{1}{2} × 4a)

0.796  = 2a

a = 0.796 / 2

a = 0.398 m/s²

using first equation of motion

V_{f} = u + at

we substitute

V_{f} = 0 + 0.398 × 2

V_{f} = 0.796 m/s

now, average velocity is given as;

V_{avg} = ( 0.796 m/s  + 0 ) / 2

V_{avg} = ( 0.796 m/s  + 0 ) / 2

now, distance as the block moves in 2s will be;

D = [( 0.796 m/s  + 0 ) / 2 ] × 2

D = 0.796 m

Therefore, distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

5 0
2 years ago
. What is the kinetic energy of a 10 kg car that is moving 4 m/s?
Darya [45]

Answer:i belive the answer is E = 1/2 mv^2

hope this helps plz mark me as braliest

8 0
2 years ago
You are participating in a NASA traineeship, working with a group planning a new landing on Mars. Your supervisor has come up wi
aivan3 [116]

Answer:

h=17005.8 km

Explanation:

Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass M=6.39\times10^{23} kg at a distance r will be:

F=\frac{GMm}{r^2}

where G=6.67\times10^{-11}Nm^2/kg^2 is the gravitational constant.

This force is the centripetal force the satellite experiments, so we can write:

F=ma_{cp}=mr\omega^2=mr(\frac{2\pi}{T})^2=\frac{4\pi^2mr}{T^2}

Putting all together:

\frac{GMm}{r^2}=\frac{4\pi^2mr}{T^2}

which means:

r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}

Which for our values is:

r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km

Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km , which leaves us with:

h=r-R=20395.3km-3389.5 km=17005.8 km

6 0
2 years ago
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