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OleMash [197]
3 years ago
14

A car is traveling around a horizontal circular track with radius r = 210 m at a constant speed v = 23 m/s as shown. The angle θ

A = 23° above the x axis, and the angle θB = 53° below the x axis.
1) What is the magnitude of the car’s acceleration?
2) What is the x component of the car’s acceleration when it is at point A
3) What is the y component of the car’s acceleration when it is at point A
4) What is the x component of the car’s acceleration when it is at point B
5) What is the y component of the car’s acceleration when it is at point B
Physics
1 answer:
Fantom [35]3 years ago
6 0

Centripetal acceleration of car is given by formula

a_c = \frac{v^2}{R}

now plug in the values in this

a_c = \frac{23^2}{210}

a_c = 2.52 m/s^2

Part b)

At position A we have

x component of acceleration is given as

a_x = -a_c cos23

a_x = -2.32 m/s^2

Part c)

At position A we have

y component of acceleration is given as

a_y = -a_c sin23

a_y = -0.98 m/s^2

Part d)

At position B we have

x component of acceleration is given as

a_x = -a_c cos53

a_x = -1.52 m/s^2

Part e)

At position B we have

Y component of acceleration is given as

a_y = a_c sin53

a_y = 2.01 m/s^2

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professor190 [17]

Answer:

A

Explanation:

I’m not sure but I hope this helped

8 0
2 years ago
How long does a car (1000 kg) have a speed of 30 m/s from a rest if the engine power is 10kw
Lemur [1.5K]

Answer: 90.1 s

Explanation:

Use equation for power:

P=F*V

Use eqation for force:

F=ma

F---force

V---velocity

Vr=om/s

V=30m/s

m=1000kg

P=10000W

---------------------------

P=FV

F=P/V

F=10000W/30m/s

F=333.33N

Use equation for force to find accelartaion.

F=ma

a=F/m

a=333.33N/1000kg

a=0.333 m/s²

Use equation for accelaration to find out time:

a=(V-Vs)/t

t=(V-Vs)/a

t=(30m/s)/(0.333m/s²)

t=90.09 s≈90.1 s

------------------------

5 0
3 years ago
A satellite with a mass of 2,000 ㎏ is inserted into an orbit that is twice the Earth' s radius. W hat is the force of gravity on
dolphi86 [110]

Answer:

option (b) 4900 N

Explanation:

m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R

F = G Me x m / (R + h)^2

F = G Me x m / 2R^2

F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2

F = 4900 N

4 0
3 years ago
A 5 kg block is being pushed horizontally across a level surface at a constant velocity. What is the magnitude of the Normal for
luda_lava [24]

Answer:

50 N.

Explanation:

On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:

\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}

The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:

\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}

In conclusion, the normal force acting on the block is 50 N.

5 0
2 years ago
An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
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