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just olya [345]

2 years ago

13

1. A block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts

from rest, and its final velocity is 12.6 m/s.

1 answer:

Snezhnost [94]2 years ago

3 0

Answer:

397 j

Explanation:

Because 5.0kg yuh

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A gnat takes off from one end of a pencil and flies around erratically for 41.641.6 s before landing on the other end of the sam

wlad13 [49]

Answer:

average speed is 0.159 m/s

average velocity = 0.0011 m/s

Explanation:

given data

time = 41.6 s

total distance = 6.65 m

length = 0.0463 m

to find out

average speed and the magnitude average velocity

solution

we know average speed formula is

average speed = $\frac{total distance}{total time}$ ...............1

put here value

average speed = $\frac{6.65}{41.6}$

average speed is 0.159 m/s

and

average velocity formula is

average velocity = $\frac{displacement}{total time}$ ...............2

here displacement is initial point to final point and here is 0.0463 m

put here value

average velocity = $\frac{0.0463}{41.6}$

average velocity = 0.0011 m/s

6 0

3 years ago

(FLUID MECHANICS)

Vera_Pavlovna [14]

Answer:

option B

Explanation:

When a body is immersed in liquid there will be two force is acting on the body.

First one force acting downward due to weight of the body.

And the second force acting on the object will be buoyant force.

If the object is not in equilibrium the apparent weight will be equal to net force acting on the object.

$F_{net} = W - F_b$

W is the weight of the object acting downward

Fb is the buoyancy force acting upward on the object.

Hence, the correct answer is option B

3 0

3 years ago

A flat circular loop of wire of radius 0.50 m that is carrying a 2.0-A current is in a uniform magnetic field of 0.30 T. What is

Luba_88 [7]

Answer:

The magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J

Explanation:

Given;

radius of the circular loop of wire = 0.5 m

current in circular loop of wire = 2 A

strength of magnetic field in the wire = 0.3 T

τ = μ x Bsinθ

where;

τ is the magnitude of the magnetic torque

μ is the dipole moment of the magnetic field

θ is the inclination angle, for a plane area perpendicular to the magnetic field, θ = 90

μ = IA

where;

I is current in circular loop of wire

A is area of the circular loop = πr² = π(0.5)² = 0.7855 m²

μ = 2 x 0.7885 = 1.571 A.m²

τ = μ x Bsinθ = 1.571 x 0.3 sin(90)

τ = 0.4713 J

Therefore, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J

4 0

3 years ago

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

You fill a car with gasoline. The car now has...

ValentinkaMS [17]

Answer:

Fuel or a provided source of energy for combustion

8 0

2 years ago