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Masteriza [31]
4 years ago
6

Which of the following numbers is equal to an object's acceleration?

Physics
2 answers:
barxatty [35]4 years ago
7 0

For the given all options we can find the respective quantity

whenever we have to relate slope of a graph then it is always derivative of quantity on Y axis with respect to X axis

while if we need to find the area of the graph then it is integral of quantity on Y axis with respect to x axis

A. The area under that object's velocity-time graph

this shows area of velocity time graph = \int v dt = displacement

so its not correct

B. The slope of that object's position-time graph

This shows slope of position time graph = \frac{ds}{dt} = velocity

so this is correct for velocity

C. The slope of that object's velocity-time graph

This shows slope of velocity time graph = \frac{dv}{dt} = acceleration

D. The area under that object's position-time graph

this shows the area of position time graph = \int  s dt

this is no physical quantity

So option B is correct

Sergio039 [100]4 years ago
4 0

Answer: C

Explanation:

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A train is moving at a speed of 50 km/h. How many hours will it take the train to travel 600 kilometers?
EastWind [94]

Answer:

12 hours

Explanation:

600 divided by 50 is 12

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3 years ago
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The momentum difference of a 3 kg mass with initial velocity of 8 m/s and a final velocity of 16 m/s
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Answer:

6

Explanation:

because if 3kg is 8m/s then 16 divided by 2 is 8 means add 3 to 3 and get the final answer

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3 years ago
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A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The
Damm [24]

Answer:

ugmd = 1/2 kx²

d = (1/2 kx²) / (ugm)

= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)

= 7.4 m

ugmd = 1/2 mv²

v = √2ugd

= √(2(0.23)(9.81 m/s²)(7.4 m)

= 5.8 m/s

Explanation:

3 0
4 years ago
Which of these will always produce a magnetic field?
vovikov84 [41]

Answer:

Technically everything has somewhat of a magnetic field. I guess

7 0
3 years ago
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
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