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OLga [1]
3 years ago
12

Two trains are traveling side-by-side along parallel, straight tracks at the same speed. In a time t, train A doubles its speed.

In the same time train B increases its speed by a factor of 5.42. By what factor is the distance traveled by train B in time t greater than the distance traveled by train A in the same time t?
Physics
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer:

derdddrdickdd

Explanation:

You might be interested in
b. A force 100N makes an angle of Ø with the x axis and has a y component of 30 N. Find both the x and y component of the force
horsena [70]

Answer:

Explanation:

The y component of the force is 100 sinØ . But given that y component is 30N

so 100 sinØ = 30

sinØ = 0.3

Ø = 17.5°.

X component of force = 100 cosØ

= 100 cos17.5

= 95.35 N .

Y component of force = 30 N .

Angle Ø = 17.5°.

6 0
3 years ago
Will give brainliest to right answer!
viva [34]
The correct answer is D.
8 0
3 years ago
Friction occurs when microscopic hills and valleys stick together<br><br> true or false
Reika [66]

Answer:

true

Explanation:

Friction occurs because no surface is perfectly smooth. Rougher surfaces have more friction between them. Heavier objects also have more friction because they press together with greater force. Friction produces heat because it causes the molecules on rubbing surfaces to move faster and have more energy.

4 0
3 years ago
Read 2 more answers
A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

8 0
3 years ago
a bus starting from rest moves with a uniform acceleration of 0.1 metre per second square for 2 minutes find the speed acquired
blondinia [14]
S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)

S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m

Distance double 720m*2=1440m

V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
6 0
3 years ago
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