PHYSICS COLLEGE

The cart is initially at rest. ForceF is applied to the cart for time ? t, after which the car has speed v. Suppose the same force is applied for the same time to a second cart with twice the mass. Friction is negligible. Afterward, the second cart's speed will be
a) 2v
b) 1/2v
c) v
d) 4v
e) 1/4v

Answers

Answer 1
Answer:

Answer:

v'=(1)/(2)v

Explanation:

Given that,

Initial speed of the cart, u = 0

Let F force is applied to the cart for time \Delta t after which the car has speed v. The force on an object is given by :

F = ma

m is the mass of the cart

We need to find the speed of second cart, if the same force is applied for the same time to a second cart with twice the mass. Force becomes,

F=(mv')/(t)

v'=(F t)/(2m)

v'=(1)/(2)v

So, the speed of second cart is half of the initial speed of first cart. So, the correct option is (b).

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an object down, but this is not true. If you place a box of mass 8 kg on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed of the belt, which is 5 m/s. The coefficient of kinetic friction between box and belt is 0.6. (a) How much time does it take for the box to reach this final speed

How can socialism
impact populations?

Answers

Answer:

it represents a fundamental difference. (more info below)

Explanation:

Production is incessantly developing and expanding in socialist countries, and employment is guaranteed for the entire productive population. Consequently, the relative overpopulation problem has been eliminated. This represents the fundamental difference between socialism's demographic law and capitalism's law.

hope this helped!
It represents a fundamental difference by gaining friends and losing friends or gaining jobs and losing jobs etc

Which of the following statements is true about the variation of pressure in function of the depth? O Pressure decreases exponentially with the depth O Pressure increases exponentially with the depth O Pressure decreases linearly with the depth o Pressure increases linearly with the depth O None of the above

Answers

Answer:

The answer is: Pressure increases linearly with the depth

Explanation:

In this case, the definition of pressure is:

P = (F)/(A)

where F = mg is the weight of the fluid over the body, and A is the area of the surface to which the force is exerted. If we consider \rho = m/V, then

P = (mg)/(A) = (\rho Vg)/(A).

Volume can be expressed as V = A*h, where A is the cross section of the column of the fluid over the body and h is the height of the column, in other words, the depth.

So

P = (A\rho gh)/(A)= \rho g h,

which means that pressure increases linearly with the depth in a factor of \rho g.

A 2.4 kg toy oscillates on a spring completes a cycle every 0.56 s. What is the frequency of this oscillation?

Answers

Answer:

f=1.79Hz

Explanation:

The period is defined as the time taken by an object to complete a cycle in a simple harmonic motion. As the frequency of the motion increases, the period decreases. Therefore, they are inversely proportional. The frequency does not depend on the mass of the object.

f=(1)/(T)\nf=(1)/(0.56 s)\nf=1.79Hz

Michelle recently started selling her invention: A bed that looks like it floats in mid-air. The bed is actually suspended by magnetic forces. Michelle is a(n)

Answers

Answer:

Explanation:

designer

illusionist

engineer

entrepreneur

salesperson

human

inventor

A thin aluminum meter stick hangs from a string attached to the 50.0 cm mark of the stick. From the 0.00 cm mark on the meter stick hangs a 5.202 kg concrete block. From the 75.0 cm mark on the meter stick hangs a 7.99 kg steel ball. At what mark on the meter stick must a 2.46 kg wooden block be attached so that the meter stick balances horizontally when lowered into fresh water? Assume the densities of concrete, steel, and wood are 2500.0 kg/m3, 8000.0 kg/m3, and 500 kg/m3 respectively. A. 57.6 cm
B. 57.4 cm
C. 57.2 cm
D. 57.8 cm

Answers

The wooden block must be attached at the 57.6 cm mark on the meter stick.

What is the value

We can solve this equation for the distance of the wooden block from the pivot point, which is the mark on the meter stick where it must be attached so that the meter stick balances horizontally when lowered into fresh water.

The density of aluminum is 2700 kg/m³ The volume of the meter stick is 0.01 m³ (length times width times thickness). The acceleration due to gravity is 9.81 m/s²

Substituting these values into the equation above, we get:

distance of wooden block from pivot point = -44.89 Nm / [(2700 kg/m³)(0.01 m³)(9.81 m/s²)]

= 0.576 m

Learn more about meter

brainly.com/question/1578784

#SPJ3

Answer:

Stop cheating in exam

Explanation:

Shame!!!!

I am sorry but I will have to refer you to the student conduct at UTA.

A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be in- dependently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes 28 s to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular

Answers

Answer:

t = 39.60 s

Explanation:

Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

            F = m a₀          

           a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

           x = v₀ t + ½ a t²

           x = ½ a₀ t²

           t² = 2x / a₀

           28² = 2x /a₀          (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction

           

in the direction of axis a, there is a motor so its force is F/2

               

the acceleration on this axis is

          a = F/2m

          a = a₀ / 2

so if we use the distance equation

             x = v₀ t + ½ a t²

as part of rest v₀ = 0

             x = ½ (a₀ / 2) t²

             

let's clear the time

             t² = (2x / a₀)  2

we substitute the let of equation 1

             t² = 28² 2

             t = 28 √2

             t = 39.60 s
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