Suppose that a 710 kg car is traveling at 13 m/s. Its brakes can apply a force of 4000 N.

What is transmitted by all waves? energy mass matter sound

disa [49]

Answer: energy

Explanation: A wave is a disturbance that transfers energy as it travels through medium.

7 0

3 years ago

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How many meters in 2.50 miles? (Use these two conversions: 1000 m = 1 km and 1.00 km = .621 mi )

Artyom0805 [142]

2.50 miles is equal to 4026 m.

<u>Explanation:</u>

As it is known that 1000 m =1 km and 1 km = 0.621 miles. So first we have to convert miles to km and then to metre as follows.

As 1 km = 0.621 miles, then

$\text { 1 miles }=\frac{1}{0.621} \mathrm{km}$

So, 2.50 miles will be equal to

$2.50 \text { miles }=\frac{2.50}{0.621} \mathrm{km}=4.026 \mathrm{km}$

Then, in order to get the answer in meters, we have to convert this km to meter by the conversion of 1000 m =1 km. So,

$1 \mathrm{km}=1000 \mathrm{m}$

Thereby,

$4.026 \mathrm{km}=4026 \mathrm{m}$

7 0

3 years ago

1. The choice of materials for an exciting playground slide should __________.

gavmur [86]

have a large coefficient of kinetic friction between cloth and the slide material
40(0.3+0.5)=40(0.8)=32N
F-applied=Wbook/μk
When the tire is locked, the frictional force is greater but the car cannot be steered

4 0

2 years ago

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Newton

vodomira [7]

Given parameters:

Mass on earth = 50kg

Unknown:

Mass on planet Xenon = ?

Weight on planet Xenon = ?

Mass is the amount of matter contained in a particular substance.

Weight is the force on a body and it is derived from the product of mass and acceleration due to gravity.

Weight = mass x acceleration due to gravity

Planet Xenon has half the gravitational force of Earth.

This translated gives $\frac{9.8}{2}$ = 4.9m/s²

Now, mass is always the same every where if the amount of matter in a substance does not change.

In this problem, mass = 50kg on planet xenon.

Weight = mass x acceleration due to gravity = 50 x 4.9 = 245N

The weight on Xenon is 245N and the mass is 50kg

4 0

3 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago