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2 years ago

When Royce was 10 years old, he had a mass of

Physics

1 answer:

coldgirl [10]2 years ago

6 0

Answer:

<em>The gravitational force between Royce and Earth would be doubled at 16 years.</em>

Explanation:

<em>"Newton's law of universal gravitation states that gravitation force between two masses is proportional to the magnitude of their masses and inverse-squared of their distance".</em>

Royce Scenario

At the age of 10 Royce's mass = 30kg

At the age of 16 Royce's mass = 60kg

From Newton's law of universal gravitation, an Increase in the mass of a body would amount to a corresponding increase in the gravitational force.

In the case of Royce, the mass double between the age of 10 and 16, so there would be an increase of the gravitation force by double.

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A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The

qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

4 0

2 years ago

A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t

Pie

Answer:

$I = \frac{mvb}{6}$

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

$L = I$ ω .... (1)

moment of inertia of rod rotating about its center of length b

$I = \frac{ mb^2}{12}$ ........ .(2)

using v = ωr

where w is angular velocity

and r is radius of rod which is equal to b

so we get 2v = ωb

ω = 2v/b ................. (3)

here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

$L = \frac{mb^2}{12}$ × $\frac{2v}{b}$

so final answer will be $L = \frac{mvb}{6}$

7 0

2 years ago

A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s

BlackZzzverrR [31]

Answer:

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

$Sf^2=So^2+(2*a*x)$

$Sf=So+(a*t)$

We are going to put our data:

$(15m/s)^2=So^2+(2*a*60m)$

$15m/s=So+(a*6s)$

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

$\sqrt{(15m/s)^2-(2*a*60m)}=So$

$15m/s-(a*6s)=So$

$\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)$

$[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}$

$(15m/s)^2-(2*a*60m)}=(15m/s)^{2}-2*(a*6s)*(15m/s)+(a*6s)^{2}$

$-120m*a=-180m*a+36s^{2}*a^{2}$

$0=120m*a-180m*a+36s^{2}*a^{2}$

$0=-60m*a+36s^{2}*a^{2}$

$0=(-60m+36s^{2}*a)*a$

$0=a1$

$\frac{60m}{36s^{2}} = a2$

$1.67m/s^{2}=a2$

If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

$Sf=So+(a*t)$

$15m/s=So+1.67m/s^2*6s$

$15m/s-(1.67m/s^2*6s)=So$

$4.98m/s=So$

For part c:

We are going to use:

$Sf^2=So^2+(2*a*x)$

$(4.98m/s)^2=0^2+(2*(1.67m/s^2)*x)$

$\frac{24.80m^2/s^2}{3.34m/s^2}=x$

$7.42m=x$

5 0

2 years ago

Dentists' chairs are examples of hydraulic-lift systems. If a chair weighs 1400 N and rests on a piston with a cross-sectional a

NeX [460]

Answer:

Force applied to smaller cross section is

= 82.63 N

Explanation:

As we know

$F_2 A_1 = F_1 A_2$

where $F1, F2$ signifies the weight of the two chair in a hydraulic-lift system

And $A_1, A_2$ signifies the area of the two respective chairs in a hydraulic-lift system

Given -

$F2=1400$ N

$A1 =1220$ Square centimeter

$A_2 = 72$ Square centimeter

Substituting the given values in above equation, we get -

$1400 * 72 = F1 * 1220\\F2 = 82.63$

Force applied to smaller cross section is

= 82.63 N

8 0

2 years ago

If you lived on Saturn, which planets would exhibit retrograde motion like that observed for Mars from Earth? (Select all that a

liberstina [14]

Answer:

a) Earth

b) Mercury

c) Neptune

Explanation:

All the planets move around the sun in eastward direction, but few planet have retrograde rotation i.e in westward direction. Retrograde motion is just an apparent change in the movement of planet which means it only seems as if the planet are rotating in opposite direction. Retrograde movement of planet like Saturn, Jupiter and mars is not real. Hence, if a person lives on Saturn, then following planets will exhibit retrograde motion

a) Earth

b) Mercury

c) Neptune

4 0

2 years ago