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2 years ago

What is the average speed of a car that travels 60 meters in 2 seconds?

Physics

1 answer:

blondinia [14]2 years ago

6 0

Answer:

30 m/s

Explanation:

Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.

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A 13 kg hanging sculpture is suspended by a 95-cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at its fundame

Artyom0805 [142]

Answer:

$f=81.96 \ Hz$

Explanation:

Givens

$L=95cm$

$m_{sculpture} =13kg$

$m_{wire}=5g$

The frequency is defined by

$f=\frac{v}{\lambda}$

Where $v$ is the speed of the wave in the string and $\lambda$ is its wave length.

The wave length is defined as $\lambda = 2L = 2(0.95m)=1.9m$

Now, to find the speed, we need the tension of the wire and its linear mass density

$v=\sqrt{\frac{T}{\mu} }$

Where $\mu=\frac{0.005kg}{0.95m}= 5.26 \times 10^{-3}$ and the tension is defined as $T=m_{sculpture} g=13kg(9.81 m/s^{2} )=127.53N$

Replacing this value, the speed is

$v=\sqrt{\frac{127.53N}{5.26 \times 10^{-3} } }=155.71 m/s$

Then, we replace the speed and the wave length in the first equation

$f=\frac{v}{\lambda}\\f=\frac{155.71 m/s}{1.9m}\\ f=81.96Hz$

Therefore, the frequency is $f=81.96 \ Hz$

3 0

2 years ago

Read 2 more answers

A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60

Bumek [7]

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm

5 0

2 years ago

Read 2 more answers

Lightwaves travel from the air into a lens made of glass. Their velocity decreases as they enter the glass. How does this affect

Papessa [141]

The waves become longer but slower

4 0

3 years ago

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

Burka [1]

Solution :

a). B at the center :

$=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

$=\frac{16}{21}=\frac{I_2}{32}$

$I_2=24.38$ A

Therefore, the current in the outer wire is 24.38 ampere.

3 0

2 years ago

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a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

Where:

m is the mass of the car
v is the speed
R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0

2 years ago