All categories

1 year ago

Several students measured the motion of a remote control car. They recorded the data in the following chart.

Physics

1 answer:

Rainbow [258]1 year ago

5 0

A)At 5.5sec car is at rest because the distance is same as at 5.0sec .

B) At t= 2sec where the distance covered by a car is 20m.

The speed of the car is 10m/s

C) Average speed

4/Savg= 0.14+6.4+5.8+5p.27

Savg = 0.227 m/s

What is non uniform motion ?

When the object is moving with unequal distance along with unequal time intervals.

to learn more about Motion click brainly.com/question/18345181

#SPJ9

You might be interested in

The slope at point A of the graph given below is:<br><br><br> WILL MARK BRAINLIEST TO CORRECT ANSWER

steposvetlana [31]

RQ/PQ I think

rise/run

3 0

3 years ago

The cross sectional area of an optical fibre is 6.3 x 10^-6 m^2 The intensity of the light entering the optical fibre is 3.2 x 1

gogolik [260]

the answer is 0.0. also known as a emoji face!

7 0

3 years ago

You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of

solong [7]

Answer:

The velocity of the man from the frame of reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

V₂₁ = V₂ -V₁

∴ V₂ = V₂₁ + V₁

On substitution

V₂ = 3 + 2

= 5 m/s

Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s

8 0

3 years ago

Help with question 16 and 17

storchak [24]

Your answer is correct.

___

The fractional change in resistance is equal to the given temperature coefficient multiplied by the change in temperature.

R = R₀×(1 + α×ΔT)

R = (10.0 Ω)×(1 + 0.004×(65 -20)) = 11.8 Ω

5 0

4 years ago

Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The on

astraxan [27]

Answer: (a) $F=7(10)^{-7}N$

(b) $F=1.344(10)^{-6}N$

(c) The force of Jupiter on the baby is slightly greater than the the force of the father on the baby.

Explanation:

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

Knowing this, let's begin with the answers:

<h2 /><h2>(a) Gravitational force Father exertes on baby</h2>

Using equation (1) and taking into account the mass of the father $m_{1}=100kg$ , the mass of the baby $m_{2}=4.20kg$ and the distance between them $r=0.2m$ , the force $F_{F}$ exerted by the father is:

$F_{F}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(100kg)(4.20kg)}{(0.2m)^2}$ (2)

$F_{F}=0.0000007N=7(10)^{-7}N$ (3)

<h2>(b) Gravitational force Jupiter exertes on baby</h2>

Using again equation (1) but this time taking into account the mass of Jupiter $m_{J}=1.898(10)^{27}kg$ , the mass of the baby $m_{2}=4.20kg$ and the distance between Jupiter and Earth (where the baby is) $r_{E}=6.29(10)^{11}m$ , the force $F_{J}$ exerted by the Jupiter is:

$F_{J}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.898(10)^{27}kg)(4.20kg)}{(6.29(10)^{11}m)^2}$ (4)

$F_{J}=0.000001344N=1.344(10)^{-6}N$ (5)

<h2>(c) Comparison</h2>

Now, comparing both forces:

$F_{J}=0.000001344N=1.344(10)^{-6}N$ and $F_{F}=0.0000007N=7(10)^{-7}N$ we can see $F_{J}$ is greater than $F_{F}$ . However, the difference is quite small as well as the force exerted on the baby.

7 0

3 years ago