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katovenus [111]

3 years ago

7

The lower the value of the coefficient of friction, the___ the resistance to sliding.

1 answer:

Nuetrik [128]3 years ago

4 0

The lower the value of the coefficient of friction, the lower the resistance to sliding.

<u>Explanation:</u>

The coefficient of friction defines as directly proportionate with the resisting force, which is the frictional force. Hence, if there seems a decrease at coefficient of friction, then it is sure that the frictional force decreases.

We know that the frictional force on a body, is the product of coefficient of frictions and the normal forces acting on the body. Note that friction acts only, if a body is in contact, and it is of three types, static, kinetic and rolling.

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Which equation relates charge, time, and current?

Ierofanga [76]

Answer:

I = Δq / t

Explanation:

The quantity of electricity i.e charge is related to current and time according to the equation equation:

Q = It

Δq = It

Where:

Q => is the quantity of electricity i.e charge

I => is the current.

t => is the time.

Thus, we can rearrange the above expression to make 'I' the subject. This is illustrated below:

Δq = It

Divide both side by t

I = Δq / t

6 0

2 years ago

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert

cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, $k=4.2\times 10^5\ N/m$

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

$mgh=\dfrac{1}{2}kx^2$

x is the compression in spring

$x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m$

So, the compression in the spring is 5.88 meters.

6 0

3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

Nadusha1986 [10]

Answer:

The charge resides on the outer surface = $1.245 \times 10^{-12}$ C

Explanation:

Surface area of cell $(A) = 4.3\times 10^{-9} m^{2}$

Separation between two plate $(d) = 1.1 \times 10^{-8} m$

Dielectric constant $(k) = 4.2$

Potential difference $(\Delta V) = 85.7 \times 10^{-3} V$

The capacitance of parallel plate capacitor in free space is given by,

$C = \frac{\epsilon_{o} A }{d}$

Where $\epsilon_{o} =$ permittivity of free space = $8.85 \times 10^{-12}$

The Capacitance of capacitor is increase by $k$ times when it placed in dielectric medium.

$C_{dielectric} = \frac{k \epsilon_{o} A }{d}$

And we know that, $C = \frac{Q}{ \Delta V}$

So charge on the outer surface is given by,

$Q = \frac{k \epsilon A \Delta V }{d}$

$Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3} }{1.1 \times 10^{-8} }$

$Q = 1.245 \times 10^{-12}$

3 0

3 years ago

working alongside the pharmacist , one of the duties a medication reconciliation technician would perform is to a ) deal with va

Vanyuwa [196]

Answer:

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5 0

2 years ago

2. A person began running due east and covered 15 kilometers in 2.0 hours. What is the average velocity of the person?

Angelina_Jolie [31]

Answer:

7.5 km/h (2.1 m/s) due east

Explanation:

The average velocity of the person is given by:

$v=\frac{d}{t}$

where

d is the displacement

t is the time taken

In this problem,

d = 15 km is the displacement

t = 2.0 h is the time elapsed

so the average velocity is

$v=\frac{15 km}{2.0 h}=7.5 km/h$

and the direction is the same as the displacement (east).

We can also convert the velocity into SI units (m/s). We have:

d = 15 km = 15,000 m

t = 2.0 h * 3600 s/h = 7200 s

$v=\frac{15,000 m}{7200 s}=2.1 m/s$

4 0

3 years ago

Read 2 more answers