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2 years ago

11

a car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achiev

e this speed?

1 answer:

atroni [7]2 years ago

6 0

T=Vf-Vi/s
25m/s -15m/s/ 125m
10m/s /125m
=0.08s
I hope it’s correct !

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Luz, who is skydiving, is traveling at terminal velocity when she changes position by going headfirst toward the ground

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How fast is she going

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3 years ago

Read 2 more answers

A clay ball with a mass of 0.48kg has an initial speed of 4.08m/s it strikes a 3.04kg clay ball at rest and the two balls stick

Rom4ik [11]

Final velocity = 0, thus final kinetic energy is 0
Initial kinetic energy:
0.5mv²
= 0.5 x 0.48 x 4.08²
= 4.0 J
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3 years ago

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.

kherson [118]

Answer:

<em>a. The rock takes 2.02 seconds to hit the ground</em>

<em>b. The rock lands at 20,2 m from the base of the cliff</em>

Explanation:

Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.

The time taken by the object to hit the ground is calculated by:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

$\displaystyle d=v.t$

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.

a,

The time taken by the rock to reach the ground is:

$\displaystyle t=\sqrt{\frac{2*20}{9.8}}$

$\displaystyle t=\sqrt{4.0816}$

t = 2.02 s

The rock takes 2.02 seconds to hit the ground

b.

The range is calculated now:

$\displaystyle d=10\cdot 2.02$

d = 20.2 m

The rock lands at 20,2 m from the base of the cliff

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2 years ago

Individuals with favorable traits are ________ in an environment

77julia77 [94]

Succesful is correct.

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3 years ago

In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s

sladkih [1.3K]

Answer:

quality factor (Q) = 69.99

inductor = 1.591 x 10⁻⁴ H

capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R = 10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW) $= \frac{R}{2\pi L}$

$BW = \frac{R}{2\pi L}$

make L (inductor) the subject of the formula

$L = \frac{R}{2\pi *BW} = \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH$

$F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}$

make C (capacitor) the subject of the formula

$C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F$

quality factor (Q) $= \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99$

quality factor (Q) = 69.99

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3 years ago