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torisob [31]
3 years ago
8

Jake's toy car is sitting on a smooth, flat surface. This wheel and axle simple machine will remain in the same place because th

e ________ force prevents the car from rolling away on its own.
Physics
1 answer:
weeeeeb [17]3 years ago
8 0

Answer:

Inertial

Explanation:

By the Newton's first law of motion every body continues to be in the state of rest or in uniform motion until an external force is applied to it.

Here the car sits on a friction-less surface so no force of friction will act on it and the wheels will continue to rotate irrespective of the position of the car.

The toy car will continue to be in the same place until and external force acts on it and causes acceleration in it.

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Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

7 0
3 years ago
A 1kg mass is thrown to a height of 2cm. what is the potential energy​
prohojiy [21]
  • Mass=m=1kg
  • Height=h=2cm=0.02m
  • Acceleration due to gravity=g=10m/s^2

\\ \tt\hookrightarrow P.E=mgh

\\ \tt\hookrightarrow PE=1(10)(0.02)

\\ \tt\hookrightarrow PE=0.2J

4 0
2 years ago
Read 2 more answers
A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250
Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block m=0.5 kg

inclination \theta =30

coefficient of static friction \mu =0.35

coefficient of kinetic friction \mu _k=0.25

distance traveled d=77.3 cm

spring constant k=35 N/m

work done by gravity+work done by friction=Energy stored in Spring

mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}

mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}

0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}

x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}

x=0.234 m

x=23.4 cm

6 0
4 years ago
Matter comprises all of them and among them are independent. what are they
telo118 [61]

<em>Matter is composed of elementary particles i.e. quarks and leptons.</em>

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8 0
3 years ago
A train moves from rest to a speed at 25m/s in 30.0 seconds. What is it’s acceleration?
sasho [114]

Answer:

25/30 = 5/6 m/s^2 5/6 meters per second squared

4 0
3 years ago
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