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AnnZ [28]
2 years ago
8

A truck of mass 8000 kg is moving with a velocity of 50 km/h and collides with a bus of mass 5000 kg coming from opposite direct

ion with a velocity of 45 km/h. After collision they stacked together and started to move in combined.
a. Define friction.
b. Why walking is difficult in a muddy street?
c. On the basis of stem calculate the combine velocity and direction.
d. Does the collision elastic? Explain mathematically.​
Physics
1 answer:
liraira [26]2 years ago
6 0

Answer:

A: 456

B: Because the shoes press down on the mud and get taken in

C:643

D:Because e = mc2 divided by 3 equals a = 3 equals 4.321a

Explanation:

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A second-grade teacher dropped a box of paper clips and they scattered all over the floor. She then asked her students, "Why wil
Yuliya22 [10]

The magnet will be a useful tool to pick the paper clips because the magnet can attract the paper clips.

<h3>Attraction of magnets</h3>

The like poles of magnets repel while unlike poles of magnets attracts. Magnets attracts irons or metallic materials.

The paper clips are mettalic or made of iron and hence the magnet will attract them.

There we can conclude that the magnet will be a useful tool to pick the paper clips because the magnet can attract the paper clips and will help to gather them together for easy picking.

Learn more about magnets here: brainly.com/question/14997726

8 0
1 year ago
The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC
enyata [817]

Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

Explanation:

<u>Given:</u>

  • Electric field in the region, \vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.
  • Charge placed into the region, q = 10.4\ nC = 10.4\times 10^{-9}\ C.

where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

Thus, the y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

3 0
2 years ago
Hope you all doing okay
Vsevolod [243]

Answer:Thank you

Explanation:

8 0
1 year ago
A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th
Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

E=-\frac{d}{dt}(BACOS\alpha )\\

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using \pi r^{2} \\ where r is the radius of 5cm or 0.05m

A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\

since we no that the angle is at 0^{0}

we determine the magnitude of the magnetic filed

B=0.5e^{-t} \\t=2s

E=-(0.5e^{-2} * 0.00785)

E=-0.000532v\\

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A

I=2.6mA

6 0
3 years ago
A block pushed along the floor with velocity V0 slides a distance d after the pushing force is removed. a) if the mass of the bl
KengaRu [80]

Answer:

a) \ d_2=d_1\\b) \ d_2=4d_1

Explanation:

Assume that the distance travelled initially is d.

In order to stop the block you need some external force which is friction.

If we use the law of energy conservation:

E_i=E_f\\\frac{mv^2}{2}= E_{Friction}\\E_{Friction}=F_{Friction}*d\\F_{Friction}= \mu_kmg\\\frac{mv^2}{2}= \mu_kmgd\\ d=\frac{v^2}{2\mu_kg}

a)

Looking at the formula you can see that the mass doesn't affect the distance travelled, as lng as the initial velocity is constant (Which indicates that the force must be higher to push the block to the same speed) therefore the distance is the same.

b) If the velocity is doubled, then the distance travelled is multiplied by 4, because the distance deppends on the square of the velocity.

6 0
3 years ago
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