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3 years ago

Exercising with a friend or partner will enhance your safety. true or false

Physics

2 answers:

bagirrra123 [75]3 years ago

7 0

Exercising with a friend or partner will enhance your safety.

true

Diano4ka-milaya [45]3 years ago

6 0

Yes that is True because if you get injured and your fried Is there he or she can get help

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A car travels at the speed of 117km/h.How far will the car travels in 50 minutes

lorasvet [3.4K]

Answer:

117/60*50

Explanation:

Trust me bc I'm smart

8 0

3 years ago

Read 2 more answers

A 5.00-kg box slides 5.50 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floo

Stels [109]

Answer:

μk = 0.08

Explanation:

Applying the work-energy theorem, we know that the change in kinetic energy of the box, is equal to the total work done on it.
The only force acting on the box capable of doing work, is the dynamic friction force.
This force is just the product of the coefficient of kinetic friction, and the normal force.
In this case, if the floor is level, the normal force is equal and opposite to the gravity force, Fg.
So we can write the following expression:

$\Delta K = -\frac{1}{2} * m* v_{0} ^{2} = \mu_{k} *m*g* cos (180\º)$

Replacing by the givens, we can solve for μk, as follows:

$\mu_{k} = \frac{v_{0}^{2} }{2*g*d} = \frac{9 (m/s)2}{2*9.8 m/s2*5.50m} = 0.08$

5 0

3 years ago

11 An unstretched spring is 12,0 cm long. A load of 5.0N stretches it to 15.0cm. How long will it be under a load

Rom4ik [11]

Answer: Approximately 22 cm

=========================================================

Explanation:

The unstretched spring is 12.0 cm long. When adding a load of 5.0 N, it stretches to 15.0 cm. This is a displacement of 15.0 - 12.0 = 3.0 cm, which is the amount the spring is stretched.

Convert this displacement to meters (so that it fits with the meters unit buried in Newtons).

3.0 cm = (3.0)/100 = 0.03 m

Apply Hooke's Law to find the spring constant k

F = -kx

5.0 = -k*(0.03)

k = -(5.0)/(0.03)

k = -166.667 approximately

Now we must find the displacement x when F = 15 newtons

F = -kx

-kx = F

x = F/(-k)

x = -F/k

x = -15/(-166.667)

x = 0.089 approximately

x = 0.1

The displacement to one decimal place is about 0.1 meters, which converts to 100*0.1 = 10 cm

So the spring will be stretched to about 12cm+10cm = 22 cm

4 0

3 years ago

A block is projected up a frictionless inclined plane with initial speed v0 = 7.14 m/s. The angle of incline is θ = 36.5°. (a) H

Wewaii [24]

Answer:

$(a)x=4.37m\\\\(b)t=1.225s\\\\(c) v_{f}=7.14m/s$

Explanation:

Given data

$v_{o}=7.14m/s\\\alpha =36.5^o$

For Part (a)

Starting with the -ve acceleration of the body (opposite to the gravitational force)

$a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2$

Using equation of motion

$v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m$

For Part (b)

Using the result in Part (a) we can substitute in other equation of motion to get time t:

$x=\frac{1}{2}vt\\ 4.37m=\frac{1}{2}(7.14m/s)t\\ (7.14m/s)t=2*(4.37)\\t=8.744/7.14\\t=1.225s$

For Part (c)

At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)

$a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s$