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3 years ago

How is coulomb's law similar to newton's law of gravity?

1 answer:

4 0

<span>Newton's law of gravitation is attractive, whereas Coulomb's law is attractive or repulsive. Both are proportional to the inverse square of distance.</span>

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When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees

dolphi86 [110]

Answer:

a) 1.725*10^5 N

b) 3.83*10^3 N

c) i) 173.24 kN

c) ii) 4.57 kN

Explanation:

See the attachment for calculations

6 0

3 years ago

A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

3 years ago

A drunken sailor stumbles 550 meters north, 500 meters northeast, then 450 meters northwest. What is the total displacement and

cluponka [151]

Answer:

Resultant displacement = 1222.3 m

Angle is 88.3 degree from +X axis.

Explanation:

A = 550 m north

B = 500 m north east

C = 450 m north west

Write in the vector form

A = 550 j

B = 500 (cos 45 i + sin 45 j ) = 353.6 i + 353.6 j

C = 450 ( - cos 45 i + sin 45 j ) = - 318.2 i + 318.2 j

Net displacement is given by

R = (353.6 - 318.2) i + (550 + 353.6 + 318.2) j

R = 35.4 i + 1221.8 j

The magnitude is

$R = \sqrt{35.4^{2}+1221.8^{2}}R = 1222.3 m$

The direction is given by

$tan\theta =\frac{1221.8}{35.4}\\\\\theta = 88.3^{o}$

3 0

3 years ago

The horizontal surface on which the block (mass 2.0 kg) slides is frictionless. The speed of the block before it touches the spr

ch4aika [34]

Answer:3.67 m/s

Explanation:

mass of block(m)=2 kg

Velocity of block=6 m/s

spring constant(k)=2 KN/m

Spring compression x=15 cm

Conserving Energy

energy lost by block =Gain in potential energy in spring

$\frac{m(v_1^2-v_2^2)}{2}=\frac{kx^2}{2}$

$2\left [ 6^2-v_2^2\right ]=2\times 10^3\times \left [ 0.15\right ]^2$

$v_2=3.67 m/s$

7 0

3 years ago

When the mass of the bottle is 0.125 kg, the KE is______ kg m2/s2.

DiKsa [7]

Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2

Explanation:

(1) Given mass = 0.125 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2

(2) Given mass = 0.250 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2

(3) Given mass = 0.375 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2

(4) Given mass = 0.500 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2

3 0

3 years ago