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3 years ago

5

A satellite orbiting Mars (6.39x1023kg) experiences a gravitational field of 2.40N/kg. How far from Mars is the satellite orbiti

ng?

1 answer:

Ann [662]3 years ago

4 0

Answer:

sorry dont know

Explanation:

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A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

3 years ago

emperature is most closely related to which property of a liquid? (1 point) Select one: a. the volume of the liquid b. the numbe

Rainbow [258]

Answer

b. the number of atoms in each molecule.

Explanation:

5 0

3 years ago

11. Explain why in terms of gravity and air resistance and the 2nd law, objects in free fall regardless of mass hit at the same

Alchen [17]

Answer:

When an object is dropped, it accelerates toward the center of Earth. Newton's second law states that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force.

7 0

3 years ago

Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as

viva [34]

Answer:

a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s

Explanation:

It is angular momentum given by

L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

${array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]$

Let's apply this relationship to our case

Let's start by breaking down the speed

v₀ₓ = v₀ cosn 45

voy =v₀ sin 45

v₀ₓ = 9 cos 45

voy = 9 without 45

v₀ₓ = 6.36 m / s

voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

vfy² = voy²- 2 g y

y = voy² / 2g

y = (6.36)²/2 9.8

y = 2.06 m

Let's calculate the angular momentum

L= $\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]$

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^ Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

R = vo² sin 2θ / g

R = 9² sin (2 45) /9.8

R = 8.26 m

L = $\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]$

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^ kg m² /s

5 0

3 years ago

What is electropower

OLEGan [10]

Answer:

electricity

Explanation:

8 0

3 years ago

Read 2 more answers