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Art [367]
3 years ago
14

The force of gravity pulls down on your school with a total force of 400,000 newtons. The force of gravity pulling down on your

school would be exactly twice as much if your school: a Had twice as much mass b Was twice as tall c Had twice as much volume d Covered twice as much area pleaseeee help
Physics
1 answer:
-BARSIC- [3]3 years ago
3 0

Answer:

A

Explanation:

Force of gravity can be expressed as:

F = GMM/R^2

Where

F = gravitational force

M = mass

R = distance

Also, W = mg

From the above equation, we can deduce that force of gravity actually depends on mass and not volume.

The correct answer is therefore A.

Had twice as much mass

Therefore, the force of gravity pulls down on your school with a total force of 400,000 newtons. The force of gravity pulling down on your school would be exactly twice as much if your school Had twice as much mass

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Verify that for values of n less than 8, the system goes to a stable equilibrium, but as n passes 8, the equilibrium point becom
irakobra [83]

Answer:

Biological system is one of the major causes of oscillation due to sensitive negative feedback loops. For instance, imagine a father teaching his son how to drive, the teen is trying to keep the car in the centre lane and his father tell him to go right or go left as the case may be. This is a example of a negative feedback loop of a biological system. If the father's sensitivity to the car's position on the road is reasonable, the car will travel in a fairly straight line down the centre of the road. On the other hand, what happens if the father raise his voice at the son "go right" or when the car drifts a bit to the left? The startled the son will over correct, taking the car too far to the right. The father will then starts yelling "go left" then the boy will over correct again and the car will definitely oscillate back and forth. A scenario that indicates the behavior of a car driver under a very steep feedback control mechanism. Since the driver over corrects in each direction. Therefore causes oscillations.

Explanation:

5 0
2 years ago
Compare and contrast camera obscura with what you know about modern digital photography, including cell phones.
Sonja [21]

Answer:

It captures images but does not preserve them.

5 0
2 years ago
Read 2 more answers
The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.
Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let \rho_{m} be the density of metal and \rho_{w} be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

W =0.10400\ N

mg=0.10400

\rho V g=0.10400

Vg=\dfrac{0.10400}{\rho_{m}}.....(I)

The weight of metal in water is

Using buoyancy force

F_{b}=0.10400-0.08400

F_{b}=0.02\ N

We know that,

F_{b}=\rho_{w} V g....(I)

Put the value of F_{b} in equation (I)

\rho_{w} Vg=0.02

Put the value of Vg in equation (II)

\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02

1000\times\dfrac{0.10400}{0.02}=\rho_{m}

\rho_{m}=5200\ kg/m^3

Hence, The density of the metal is 5200 kg/m³.

6 0
3 years ago
A billiards ball B rests on a horizontal surface and is struck by another billiards ball A of the same mass m = 0.2 kg. Ball A i
SOVA2 [1]

Answer:

v1 = 15.90 m/s

v2 = 8.46 m/s

mechanical energy before collision = 32.4 J

mechanical energy after collision = 32.433 J

Explanation:

given data

mass m = 0.2 kg

speed = 18 m/s

angle =  28°

to find out

final velocity and  mechanical energy both before and after the collision

solution

we know that conservation of momentum remain same so in x direction

mv = mv1 cosθ + mv2cosθ

put here value

0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)

3.6 =  0.1765 V1 + 0.09389 v2    ................1

and

in y axis

mv = mv1 sinθ - mv2sinθ

0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)

0 = 0.09389 v1 - 0.1768 v2   .......................2

from equation 1 and 2

v1 = 15.90 m/s

v2 = 8.46 m/s

so

mechanical energy  before collision = 1/2 mv1² + 1/2 mv2²

mechanical energy before collision = 1/2 (0.2)(18)² + 0

mechanical energy before collision = 32.4 J

and

mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²

mechanical energy after collision = 32.433 J

7 0
2 years ago
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit
creativ13 [48]

Answer:

Explanation:

From the given information:

Distance d_i = 4.8 \times 10^{10} \ m

Speed of the comet V_i = 9.1 \times 10^{4} \ m/s

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where;

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G = 6.67 \times 10^{-11}

To find the speed V_f:

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E_f = E_i + 0 \\ \\  K_f + U_f = K_i + U_i  \\ \\ = \dfrac{1}{2}mV_f^2 +  \dfrac{-GMm}{d^2} =  \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [  \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}

V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [  \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}

\mathbf{V_f =53.125 \times 10^4 \ m/s}

3 0
2 years ago
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