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melamori03 [73]
2 years ago
5

Which is not true of friction? A. Causes wear and tear of the surfaces B. Helps us to fall easily on roads C. Produces heat D. P

roduces light E. Stops bodies when they move over each other.
Physics
2 answers:
Ganezh [65]2 years ago
5 0

Answer:

B.

Hope this helps

Aleks04 [339]2 years ago
3 0
The answer is B

i need 20 works
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A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo
WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

acceleration=\frac{30}{5}=6m/s^{2}

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0
3 years ago
Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 130-kg raft that was initially at
Ghella [55]

Answer:u=4.04 m/s

Explanation:

Given

Mass m=85 kg

mass of Raft M=130 kg

velocity of raft and man  v=1.6 m/s

Let initial speed of Tyrone is u

Conserving Momentum as there is no external Force

mu=(M+m)v

85\times u=(85+130)\cdot 1.6

u=\frac{215}{85}\cdot 1.6

u=2.529\cdot 1.6=4.04 m/s  

4 0
3 years ago
A 5.92 g object moving to the right at 17.1 cm/s makes an elastic head-on collision with an 11.84 g object that is initially at
Lelechka [254]

Answer:

v₁f = -5.7 cm/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

        m_{1} * v_{10} = m_{1} *v_{1f} + m_{2} * v_{2f} (1)

  • Rearranging terms, we have:

        m_{1} * (v_{10} - *v_{1f} ) = m_{2} * v_{2f} (2)

  • We also know that the collision is elastic, so total kinetic energy must be conserved , as follows:

        \Delta K = 0 \\ \\ \frac{1}{2} * m_{1} *v_{10} ^{2} = \frac{1}{2}* m_{1}  *v_{1f} ^{2}  + \frac{1}{2}* m_{2}  *v_{2f} ^{2} (3)

  • Rearranging , and simplifying common terms, we have:

        m_{1}* (v_{10} ^{2} -v_{1f} ^{2} ) = m_{2}  *v_{2f} ^{2} (4)

  • Replacing by the givens, doing some algebra and dividing (4) by (2), we find the following relationship:

        v_{10} + v_{1f} = v_{2f}

  • Replacing the expression above in (1), as m₂ = 2*m₁, we can find the value of v₁f, as follows:

       m_{1} * v_{10}  = m_{1} * v_{1f} +2*m_{1} * (v_{10} + v_{1f})\\ \\ -(m_{1} * v_{10}) = 3* m_{1} *v_{1f} \\ \\ v_{1f} = - \frac{v_{10} }{3}  = \frac{-17.1cm/s}{3} = -5.7 cm/s

7 0
3 years ago
Timmy bikes to school 3 kilometers east and 2 kilometers south. What is Timmy's final displacement?
vovikov84 [41]

Answer:

3.6km South East

Explanation:

Displacement is the shortest distance between the starting point and the ending point and the direction it is displaced in. To calculate the displacement we can use the pythagoras theorem because the 3km East and the 2km south form the two shorter sides of a right angled triangle between the starting and ending points. So, the displacement is the length C of the triangle which we can calculate as follows:

Pythagoras Theorem:

a^2+b^2=c^2

(2)^2+(3)^2=c^(2)

4+9=c^2

Square root 13 = c

c=3.6km (1dp)

The total displacement is 3.6km and is in the approximate direction of South East (because he travelled east and south).

Hope this helped!

8 0
3 years ago
If you will be rinsing your regulator after removing it from the cylinder, you must make sure that the ______ ______ is firmly i
Rudik [331]

If you will be rinsing your regulator after removing it from the cylinder, you must make sure that the dust cap is firmly in place.

<h3>What is Dust cap?</h3>

A dust cap is a gently curved dome mounted either in concave or convex orientation over the central hole of most loudspeaker diaphragms.

Thus, if you will be rinsing your regulator after removing it from the cylinder, you must make sure that the dust cap is firmly in place.

Learn more about dust cap here: brainly.com/question/10723016

#SPJ1

5 0
2 years ago
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