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Kamila [148]
2 years ago
14

At one point in the rescue operation, breakdown vehicle A is exerting a force of 4000 N and breakdown vehicle B is exerting a fo

rce of 2000 N.(i) Using a scale of 1 cm = 500 N, make a scale drawing to show the resultant force on the truck. (ii) Use your diagram to find the magnitude and direction of the resultant force on the truck.

Physics
1 answer:
lukranit [14]2 years ago
8 0

Answer:

1.) Magnitude = 5596 N

2.) Direction = 60 degrees

Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N

Let us resolve the two forces into X and Y component

Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N

Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )

= 2000 + 2828.43

= 4828.43 N

The resultant force R will be

R = sqrt ( X^2 + Y^2 )

Substitutes the forces at X component and Y component into the formula

R = sqrt ( 2828.43^2 + 4828.43^2 )

R = sqrt ( 31313752.53 )

R = 5595.87 N

The direction will be

Tan Ø = Y/X

Substitute Y and X into the formula

Tan Ø = 4828.43 / 2828.43

Tan Ø = 1.707106

Ø = tan^-1( 1.707106 )

Ø = 59.64 degree

Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.

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ArbitrLikvidat [17]

Answer:

<em>c. The astronaut does not need to worry: the charge will remain on the outside surface.</em>

<em></em>

Explanation:

The astronaut need not worry because <em>according to Gauss's law of electrostatic, a hollow charged surface will have a net zero charge on the inside.</em> This is the case of a Gauss surface, and all the charges stay on the surface of the metal chamber. This same principle explains why passengers are safe from electrostatic charges, in an enclosed aircraft, high up in the atmosphere; all the charges stay on the surface of the aircraft.

3 0
2 years ago
A cord is wrapped around the rim of a solid uniform wheel 0.280m in radius and of mass 8.80kg. A steady horizontal pull of 32N t
Grace [21]

Answer:25.97 rad/s^2

Explanation:

Given

radius of wheel r=0.28 m

mass of wheel m=8.80 kg

Force F=32 N

Moment of Inertia of solid wheel I=\frac{mr^2}{2}

I=\frac{8.8\times 0.28^2}{2}

I=0.344 kg-m^2

Torque is given by

\tau =F\times r=I\times \alpha

32\times 0.28=0.344\times \alpha

\alpha =25.97 rad/s^2

Force on the axle is 32 N since there is no linear acceleration of the system

using Third law F=32 N

Torque of the axle applied to the wheel is zero because force of axle imparted at the center of axle

3 0
2 years ago
A radar used to detect the presence of aircraft receives a pulse that has reflected off an object 5 ✕ 10−5 s after it was transm
Sunny_sXe [5.5K]

Answer:

7500 m

Explanation:

The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.

We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so

d = 2L

In a time of

t=5\cdot 10^{-5}s

Electromagnetic waves travel in a vacuum at the speed of light, which is equal to

c=3.0\cdot 10^8 m/s

Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:

d=vt (1)

where

v=c=3.0\cdot 10^8 m/s

t=5\cdot 10^{-5}s

d=2L, where L is the distance between the radar and the object

Re-arranging eq(1) and substituting, we find L:

L=\frac{vt}{2}=\frac{(3.0\cdot 10^8)(5\cdot 10^{-5})}{2}=7500 m

7 0
2 years ago
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
2 years ago
The speed of sound in dry air at 20 °C is 343.5 m s-1, and the frequency of the sound from the note C# above middle C on the pia
Nastasia [14]

Answer:

1.23917 m

0.14323 s

Explanation:

v = Speed of sound in dry air at 20 °C = 343.5 m/s

f = Frequency of note C# = 277.2 /s = 277.2 Hz

λ = Wavelength

v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343.5}{277.2}\\\Rightarrow \lambda=1.23917\ m

Wavelength = 1.23917 m

Distance the wave needs to travel is 49.2 m

Time = Distance / Speed

\text{Time}=\frac{49.2}{343.5}=0.14323\ s

Time taken for the sound to travel across the concert hall is 0.14323 s

4 0
2 years ago
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