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grin007 [14]
3 years ago
12

1. Convert 100 m, the length of a well-known track event, to kilometers.

Physics
1 answer:
BartSMP [9]3 years ago
3 0

Answer:

0.1 km

Explanation:

Each km is 1000 m. 100 m is 1/10 of 1000 m. in decimal form that is 0.1 km.

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A vertical spring with stiffness k originally is at rest with no mass attached. Then, a mass M is attached, and the spring rocks
Alina [70]

Answer:

The position of the spring in terms of g, m & k is x = \frac{m g}{k}

Explanation:

Stiffness of the spring = k

Mass = m

When a mass m is attached with the spring then spring stretched. in that case the force exerted on the spring is equal to weight of the mass attached.

⇒ Force exerted on the spring F = k x

⇒ m g = k x

⇒ x = \frac{m g}{k}

This is the position of the spring in terms of g, m & k.

8 0
3 years ago
A car traveling at a speed of 50.0 m/s encounters an emergency and comes to a complete stop. How much time will it take for the
ludmilkaskok [199]

Answer:

t=1.25s

Explanation:

The formula is

a= (V1-V0) /t

t = (V1-V0)/a

V1=50m/s

V0= 0 m/s

a= - 4m/s2

t= (0-50)/-4 = 1.25s

6 0
2 years ago
The white ring seen here, within our solar system, is a(n) ________ belt.
Gelneren [198K]
Asteroid belt :-) hope it helped
3 0
3 years ago
Read 2 more answers
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
2 years ago
Comparison between copper properties and aluminium properties​
mixas84 [53]
Hopes this helps:

Answer: Aluminum has 61 percent of the conductivity of copper, but has only 30 percent of the weight of copper. That means that a bare wire of aluminum weights half as much as a bare wire of copper that has the same electrical resistance. Aluminum is generally more inexpensive when compared to copper conductors.
5 0
2 years ago
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