Where would the normal force exerted on the rover when it rests on the surface of the planet be greater

Answers

Answer 1

Answer:

Answer:

Normal force exerted on the rover would be greater at a point on the surface of the planet where the weight of the rover is experienced to be greater.

Explanation:

Since weight is a vector quantity, it can vary with position. Weight is the amount of force the planet exerts on the rover centered towards the planet.

Such a force is the result of gravitational pull and is quantified as:

$F=G* (M.m)/(R^2)$

and $M=\rho* (4\pi.r^3)/(3)$

where:

R = distance between the center of mass of the two bodies (here planet & rover)

G = universal gravitational constant

M = mass of the planet

m = mass of the rover

This gravitational pull varies from place to place on the planet because the planet may not be perfectly spherical so the distance R varies from place to place and also the density of the planet may not be uniform hence there is variation in weight.

Weight is basically a force that a mass on the surface of the planet experiences.

According to Newton's third law the there is an equal and opposite reaction force on the body (here rover) which is the normal force.

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You are exiting a highway and need to slow down on the off-ramp in order to make the curve. It is rainy and the coefficient of static friction between your tires and the road is only 0.4. If the radius of the off-ramp curve is 36 m, then to what speed do you need to slow down the car in order to make the curve without sliding?

Answers

Answer:

11.87m/s

Explanation:

To solve this problem it is necessary to apply the concepts related to frictional force and centripetal force.

The frictional force of an object is given by the equation

$F_r = \mu N$

Where,

$\mu =$ Friction Coefficient

N = Normal Force, given also as mass for acceleration gravity

In the other hand we have that centripetal force is given by,

$F_c= (mv^2)/(R)$

The force experienced to stay on the road through friction is equal to that of the centripetal force, therefore

$F_r = F_c$

$\mu mg = (mv^2)/(R)$

Re-arrange to find the velocity,

$V = √(R\mu g)$

$V = √(36*0.4*9.8)$

$V = 11.87m/s$

Therefore the speed that it is necessaty to slow down the car in order to make the curve without sliding is 11.87m/s

Which state of matter is most similar to solids

Answers

Answer:

liquids

Explanation

A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.How fast is the skier moving when she gets to the bottom ofthe hill?

Answers

Explanation:

The given data is as follows.

Mass, m = 62 kg, Initial speed, $v_(1)$ = 6.90 m/s

Length of rough patch, L = 4.50 m, coefficient of friction, $\mu_(k)$ = 0.3

Height of inclined plane, h = 2.50 m

According to energy conservation equation,

(Final kinetic energy) + (Final potential energy) = Initial kinetic energy + Initial potential energy - work done by the friction

$K.E_(2) + U_(2) = K.E_(1) + U_(1) - W_(f)$

$(1)/(2)mv^(2)_(2) + U_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL$

Since, final potential energy is equal to zero. Therefore, the equation will be as follows.

$(1)/(2)mv^(2)_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL$

Cancelling the common terms in the above equation, we get

$(1)/(2)v^(2)_(2) = (1)/(2)v^(2)_(1) + gh - \mu_(k)gL$

= $(1)/(2)(6.90)^(2)_(1) + 9.8 m/s^(2) * 2.50 m - (0.3 * 9.8 * 4.50 m)$

= 36.055 - 13.23

= 22.825

$v_(2) = √(2 * 22.825)$

= 6.75 m/s

Thus, we can conclude that the skier is moving at a speed of 6.75 m/s when she gets to the bottom of the hill.

Answer:

Explanation:

mass, m = 62 kg

initial velocity, u = 6.9 m/s

length, l = 4.5 m

height, h = 2.5 m

coefficient of friction, μ = 0.3

Final kinetic energy + final potential energy = initial kinetic energy + initial potential energy + wok done by friction

Let the final velocity is v.

0.5 mv² + 0 = 0.5 mu² + μmgl + mgh

0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.5

0.5 v² = 23.805 + 13.23 + 24.5

v² = 123.07

v = 11.1 m/s

Two identical metal spheres a and b are in contact. both are initially neutral. 1.0×1012 electrons are added to sphere a, then the two spheres are separated. you may want to review ( pages 639 - 641) . part a afterward, what is the charge of sphere a?

Answers

The charge on the sphere A and sphere B after they are separated is $\boxed{ - 80\,{\text{nC}}}$ each.

Further Explanation:

Given:

The number of electrons transferred to sphere $A$ is $1.0 * {10^(12)}$ .

Concept:

The amount of charge carried by the electrons when reaches the spheres kept in contact with each other is first distributed equally on each sphere. Later as the spheres are moved away from one another, the charge on each sphere remains the same as it was when they were in contact.

The amount of charge on one electron is $- 1.6 * {10^( - 19)}\,{\text{C}}$ .

So, the amount of charge carried by the $1.0 * {10^(12)}$ electrons is given as.

$\begin{aligned}Q&= \left( {1.0 * {{10}^(12)}} \right)\left( { - 1.6 * {{10}^( - 19)}} \right)\n&= - 1.6 * {10^( - 7)}\,{\text{C}}\n\end{aligned}$

Since the charge is disturbed equally on the two sphere, so the amount of charge carried by each sphere s half of the total charge.

$\begin{aligned}{Q_A}&= \frac{{ - 1.6 * {{10}^( - 7)}}}{2}\n&= - 8 * {10^( - 8)}\,{\text{C}}\n &= - 8{\text{0}}\,{\text{nC}}\n\end{aligned}$

Thus, the amount of the charge carried by each sphere after separating from each other is $\boxed{ - 80\,{\text{nC}}}$ .

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Answer Details:

Grade: College

Chapter: Electrostatics

Subject: Physics

Keywords: Metal spheres, two identical, in contact, neutral, charged, electrons, charge on electron, charge on metallic sphere, charge of sphere A.

The charge on the sphere A and sphere B after they are separated is $-80\mu C$ each

What is Charge?

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field.

The amount of charge carried by the electrons when reaches the spheres kept in contact with each other is first distributed equally on each sphere. Later as the spheres are moved away from one another, the charge on each sphere remains the same as it was when they were in contact.

The amount of charge on one electron is $-1.6* 10^(-19) \ C$

So, the amount of charge carried by the electrons is given as.

$Q=(1* 10^(12))(-1.6* 10^(-19))$

$Q=-1.6* 10^(-7)\ C$

Since the charge is disturbed equally on the two sphere, so the amount of charge carried by each sphere s half of the total charge.

$Q_A=(-1.6* 10^(-7))/(2)$

$Q_A=-8* 10^(-8)\ C$

$Q_A=-80\ \mu C$

Thus, the amount of the charge carried by each sphere after separating from each other is $-80\mu C$

To know more about Charge follow

brainly.com/question/25922783

What is the maximum distance allowed between the center of hole #2 and datum B as seen in the front view?

Answers

Answer:

4.003" (inches )

Explanation:

The maximum distance allowed between the center of hole #2 and datum B can be calculated by adding 4.000" + 0.003" ( perpendicularity of the of hole #2) as seen from the front view of the diagram .

Note :The hole 2 is sited below the workpiece when viewed from the front view while the Datum B is positioned on the left end of the workpiece also note that the diameter is

Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.

Answers

Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5
Frequency f = 2 Hz

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