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Agata [3.3K]
3 years ago
7

A car increases its speed from 20km/hr to 50km/hr in seconds. Its acceleration is _____.

Physics
1 answer:
mash [69]3 years ago
7 0

Answer:

Option D

0.83 m/s2

Explanation:

Time is assumed as 10 seconds

First, convert the speeds from km/h to m/s

20 km/h\times \frac {1000 m}{3600 s}=5.555555556 m/s \approx 5.56 m/s

50 km/h\times \frac {1000 m}{3600 s}=13.88888889  m/s \approx 13.89 m/s

Acceleration, a=\frac {v-u}{t} where u and v are the initial and final velocities respectively, t is the time taken to accelerate.

Substituting 13.89 m/s for v, 5.56 m/s for u and 10 s for t then

a=\frac {13.89-5.56}{10}=0.8333333 m/s^{2}\approx 0.83 m/s^{2}

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A student has a displacement of 739 m north in 162 s. What was the student’s average velocity?A. 0.22 m/sB. 119,718 m/sC. 162 m/
user100 [1]

Answer:

answer below

Explanation:

Displacement of the student is 739 m due North and it takes 162 s.

We need to find the student's average velocity. Using formula of velocity.

Velocity = displacement/time

v= 739/162

v= 4.56

4 0
3 years ago
3. A 900N mountain climber scales a
Umnica [9.8K]

Answer:

<h2>135,000 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 900 × 150

We have the final answer as

<h3>135,000 J</h3>

Hope this helps you

4 0
3 years ago
Since all objects are ‘weightless’ for an astronaut in orbit, is it possible for astronauts to tell whether an object is heavy o
ivanzaharov [21]
W=gm
where g - gravitation 
m - mass
w - weight
as gravitation equals to zero, multiplying by 0 gives W=0
It is not possible to tell whether and object is heavy or light 

4 0
3 years ago
Read 2 more answers
A force of attraction would exist between
Kryger [21]

Answer:

D. two positively charged objects

3 0
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

5 0
4 years ago
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