When the mallet hits the ball with an action force, the ball exerts a reaction force on the mallet, as explained by __

A teaspoonful of white dwarf material on earth would weigh

sashaice [31]

Answer:

15 tons

Explanation:

4 0

3 years ago

In a parallel circuit with 3 resistors, the output voltage at the power supply is recorded to be 5 A. The current flowing thro

Bogdan [553]

In a parallel circuit, the total power supply current is the sum of the
currents through all of the individual branches.

The output current (not the voltage) of the power supply is 5 A.
So the individual currents in the circuit branches is 5 A.

The current through the first 2 resistors is (1 + 0.5) = 1.5 A.

So the current through the 3rd resistor is (5 - 1.5) = <u>3.5 A</u>.

8 0

4 years ago

A new planet is discovered that has twice the Earth's mass and twice the Earth's radius. On the surface of this new planet a per

kirza4 [7]

Answer:

option B

Explanation:

Radius of new planet,R' = 2R

Mass of the earth, M' = 2 M

R and M is the Radius and Mass of the earth.

Weight of the person on earth = 500 N

Weight of the person in the new planet = ?

We know acceleration due to gravity is calculated by using formula

$g = \dfrac{GM}{R^2}$

now, acceleration due to gravity on the new planet

$g' = \dfrac{GM'}{R'^2}$

$g' = \dfrac{G(2M)}{(2R)^2}$

$g' =\dfrac{1}{2} \dfrac{GM}{R^2}$

$g' =\dfrac{g}{2}$

here acceleration due to gravity is half in new planet, so weight will also be half on the new planet.

Weight on the new plane is equal to $\dfrac{500}{2}$ = 250 N

correct answer is option B

5 0

3 years ago

The distance between the centers of the wheels of a motorcycle is 156 cm. The center of mass of the motorcycle, including the ri

Drupady [299]

Answer:

a = 9.86 m/s²

Explanation:

given,

distance between the centers of wheel = 156 cm

center of mass of motorcycle including rider = 77.5 cm

horizontal acceleration of motor cycle = ?

now,

The moment created by the wheels must equal the moment created by gravity.

take moment about wheel as it touches the ground, here we will take horizontal distance between them.

then, take the moment around the center of mass. Since the force on the ground from the wheels is horizontal, we need the vertical distance.

now equating both the moment

m g d = F h

d is the horizontal distance

h is the vertical distance

m g d = m a h

term of mass get eliminated

g d = a h

so,

$a = \dfrac{g\ d}{h}$

$a = \dfrac{9.8\times 0.78}{0.775}$

a = 9.86 m/s²

4 0

3 years ago

In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact

OLga [1]

Answer:

v=3.73m/s, $a=3.35m/s^{2}$

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

$K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}$

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

$K_{A}+U_{Ag}=U_{Cs}+W_{Cf}$

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}$

so we can now solve this for k, which is the constant of the spring.

$k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}$

and now we can substitute the respective data:

$k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}$

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

$K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}$

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}$

and we can solve for the Velocity in B, so we get:

$V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}$

we can now input all the data directly, so we get:

$V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}$

which yields:

$V_{B}=3.73m/s$

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

$\sum F=ma$