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mario62 [17]
3 years ago
6

a teacher pushed a 10kg desk across a floor for a distance of 5m. she exerted a horizontal force of 20n. how much work was done?

Physics
1 answer:
serious [3.7K]3 years ago
7 0
Work Done = Force x distance
Since she exerted a horizontal force of 20N over a distance of 5m, the work done is 20N x 5m which is equals to 100 joules
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Car A hits car B (initially at rest and of equal mass) frombehind while going 35 m/s. Immediately after the collision, car Bmove
kolezko [41]

Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2

kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2

Since the masses are equals:

m=ma=mb

For the known velocities, the kinetics energies result:

ki=0.5*mVa_{i}^2

ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m

kf=0.5*mbVb_{f}^2

kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m

The lost energy in the collision is the difference between the initial and final kinectic energies:

kl=ki-kf

kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m

Finally the relation between the lost and the initial kinetic energy:

kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m

kl/ki = 24/49=0.49

7 0
3 years ago
An aluminum wire having a cross-sectional area equal to 3.90 10-6 m2 carries a current of 6.00 A. The density of aluminum is 2.7
stich3 [128]

Answer:

Vd = 1.597 ×10⁻⁴ m/s

Explanation:

Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³

To find:

Drift Velocity Vd=?

Solution:

the formula is Vd = I/nqA         (n is the number of charge per unit volume)

n = No. of electron in a mole ( Avogadro's No.) / Volume

Volume = Molar mass / density   ( molar mass of Al =27 g)

V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³

n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)

n= 6.02 × 10 ²⁸

Now

Vd = (6A) / (  6.02 × 10 ²⁸ ×  1.6 × 10⁻¹⁹ C ×  3.9×10⁻⁶ m²)

Vd = 1.597 ×10⁻⁴ m/s

6 0
3 years ago
A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with unifor
Dahasolnce [82]

Answer:

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is  located on the x-axis=\lambda_1

Linear charge density of another long line which is  located on the y-axis=\lambda_2

We have to find the direction of electric field at z=a on the positive z-axis if \lambda_1 and \lambda_2 are positive.

The direction of electric field  at z=a on the positive z-axis  is positive z-direction .

Because \lambda_1 and \lambda_2 are positive and the electric field is  applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

6 0
3 years ago
Which shared property belongs in the region marked "B"?
NNADVOKAT [17]
The properti that you didnt share
4 0
3 years ago
What is mass times acceleration?
frozen [14]

Answer:

Force

Explanation:

The mass of an object is the quantity of matter it contains. It is measured in kilograms.

Acceleration is the ratio of the change in the velocity of an object to the change in time. It is measured in m/s^{2}.

When the mass of an object is multiplied with its acceleration, this gives the average force applied on the object. As force is defined as agent that can change the state of an object.

i.e     F = m × a

where F is the force, m is the mass of the object and a its acceleration.

The two major classes of force are; contact force and field force.

6 0
3 years ago
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