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Shtirlitz [24]
2 years ago
14

A car with a mass of 1400kg is being driven along the motorway at 30m/s. Calculate the kinetic energy of the car

Physics
1 answer:
Umnica [9.8K]2 years ago
3 0

Answer:

6.3 \cdot 10^5 J

Explanation:

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the car in this problem,

m = 1400 kg

v = 30 m/s

Therefore, the kinetic energy of the car is

K=\frac{1}{2}(1400)(30)^2=6.3 \cdot 10^5 J

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An antibaryon composed of two antiup quarks
Sveta_85 [38]

Answer:

(2) −1 e

Explanation:

A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.

Up, charm and top quarks have +\frac{2}{3} e charge where as down, strange and bottom quarks have -\frac{1}{3}e charge.

The antiparticle of up quark is antiup quark and has charge -\frac{2}{3}e charge.

The antiparticle of down quark is antidown quark and has charge +\frac{1}{3}e charge.

An antibaryon is composed of two anti-up quark and one anti-down quark.

Net charge of the anti-baryon is:

2\times (-\frac{2}{3} e)+1\times (+\frac{1}{3})e=-1e

Thus, antibaryon has -1e charge.

5 0
2 years ago
Process in which permanent deformation of metals occurs due to applied stress and results in breaking of bonds and then reformin
luda_lava [24]

Answer:

This process involves the motion of dislocations and is termed slip (or glide in some textbooks)

Explanation:

Plastic deformation of metals (and other crystalline materials) usually occurs by slip, which is the sliding of planes of atoms over one another by dislocation movements.

On a microscopic scale, stress causes planes of crystalline objects to leave their original position and slide over other planes into new positions, these microscopic movements manifest as a slip on a macroscopic scale. And the planes do not return back to their original position after the removal of the dislocation-causing stress.

6 0
2 years ago
a box weighing 155 N is pushed horizontally down the hall at constant velocity. the applied force is 83 n what is the coefficien
meriva

Answer:

μ = 0.535

Explanation:

On a level floor, normal force = weight.

N = W

Friction force = normal force × coefficient of friction.

F = Nμ

Substitute:

F = Wμ

83 = 155μ

μ = 0.535

Round as needed.

8 0
2 years ago
1 Ten (10) ml aqueous solutions of drug A (10% w/v) and drug B (25% w/v) are stored in two identical test tubes under identical
Reil [10]

Answer:

YOUR answer is given below:

Explanation:

3 0
1 year ago
1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational
sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = \frac{m v^{2} }{r}

==> 1.1 × 10^{19} = (9.4 × 10^{21} × v^{2} ) ÷ 1.5 × 10^{8}

==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

==> v= 0.419188323 × 10^{3} m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×10^{8} ÷ 0.419188323 × 10^{3}

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0
2 years ago
Read 2 more answers
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