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just olya [345]

3 years ago

11

A runner runs a 1,500 m race on a circular track. The runner stops 100 m from the starting point. What are the distance and disp

lacement traveled by the runner

1 answer:

viva [34]3 years ago

4 0

Answer:

distance- 1400 m

displacement- 100 m before the finish line

Explanation:

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A certain AM radio wave has a frequency of 24Hz. Given that radio waves travel at 3 m/s, what is the wavelength of this radio wa

Dahasolnce [82]

Answer:

Wavelength of AM radio wave = 0.125 meter

Explanation:

Given:

Frequency of AM radio wave = 24 Hz

Velocity of AM radio wave = 3 m/s

Find:

Wavelength of AM radio wave

Computation:

Wavelength = Velocity / Frequency

Wavelength of AM radio wave = Velocity of AM radio wave / Frequency of AM radio wave

Wavelength of AM radio wave = 3 / 24

Wavelength of AM radio wave = 0.125 meter

3 0

3 years ago

A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. Ho

Flauer [41]

Answer:

Average force = 67 mn

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum = F × Δt

(v-u)/t = F × Δt

F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

3 0

3 years ago

Decision making worksheet answer key

mario62 [17]

Answer:

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Explanation:

7ün go up to the meaning of the name to the class finish it was in the picture of the meaning is F C B you send me your pic of clans go for a walk and 10 even though I am not say that you wake me up

7 0

3 years ago

What are the 3 Newton's law of motions?

irinina [24]

Answer:

laws of motion relate an object’s motion to the forces acting on it. In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

4 0

2 years ago

A 1.00-kg object is attached by a thread of negligible mass, which passes over a pulley of negligible mass, to a 2.00-kg object.

Anna [14]

Answer:

a = 3.27 m/s²

v = 2.56 m/s

Explanation:

given,

mass A = 1 kg

mass B = 2 kg

vertical distance between them = 1 m

$F_d = mg$

$F_d = 2 \times 9.8$

$F_d = 19.6\ N$

$F_u = mg$

$F_u = 1 \times 9.8$

$F_u = 9.8\ N$

$F_{net} = 19.6 - 9.8$

$F_{net}=9.8\ N$

$F = (m_1+m_2)a$

$9.8 = (2+1)a$

a = 3.27 m/s²

The speed of the system at that moment is:

v² = u² + 2×a×s

v² = 0² + 2× 3.27 × 1

v ² = 6.54

v = 2.56 m/s

3 0

3 years ago