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nydimaria [60]
3 years ago
8

A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n

egotiate the curve if the horizontal component of the normal force on the car from the road is equal in magnitude to _____________.
Physics
1 answer:
Sholpan [36]3 years ago
5 0

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

N sin\theta = \frac{mv^2}{R}

N cos\theta = mg

so we have

v = \sqrt{Rg tan\theta}

so this is horizontal component of normal force is equal to the centripetal force on the car

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A coil lies flat on a level table top in a region where the magnetic field vector points straight up. The magnetic field suddenl
Leto [7]

Answer:

a) counterclockwise

Explanation:

The Lenz's Law establish that the direction of the induced current is in an opposite direction, in comparison with the change od the magnetic flux.

\epsilon=-\frac{d\Phi_{B}}{dt}

Specifically, if the flux decreases the induced current is in a counterclockwise direction, if increases, the direction is in a clockwise direction.

HOPE THIS HELPS!!

5 0
3 years ago
Line segment gj is a diameter of circle l. angle k measures (4x 6)°. circle l is inscribed with triangle g j k. line segment g j
choli [55]

The value of x in the given right triangle in a semicircle is determined as 21.

<h3>What is the measure of a triangle in a semicircle?</h3>

The triangle in a semicircle is always a right angle triangle.

From the figure shown, we can say that the triangle  G J K is right triangle and m<K = 90degrees.

Given that m<K = 4x + 6, we will can use the following equation to find the value of x as shown:

4x + 6 = 90

4x = 90 - 6

4x = 84

x = 21

Thus, the value of x in the given right triangle in a semicircle is determined as 21.

Learn more about right angle here: brainly.com/question/64787

#SPJ4

8 0
2 years ago
At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r
Oksana_A [137]

Answer:

<u>For proton:</u>

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

<u>For electron:</u>

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s    

Explanation:

<u>For a proton we have</u>:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.                

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

W = \Delta Vq    (1)                                        

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C    

V: is the potential    

Also, the work is equal to:

W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}     (2)      

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

v_{a}: is the velocity in the point a

v_{b}: is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

\Delta Vq = \frac{1}{2}mv_{a}^{2}

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}

v_{a} = 2.9 \cdot 10^{4} m/s

<u>For an electron we have</u>:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}

Since v_{a} = 0 (starts from rest) and m_{e} = 9.1x10⁻³¹ kg (electron's mass), we have:

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}    

v_{b} = 1.2 \cdot 10^{6} m/s

I hope it helps you!

6 0
4 years ago
People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th
Whitepunk [10]

Answer:

Current needed = 704A

Explanation:

Using the fomula; torque(τ) = (I)(A)(B)Sinθ

Where B = uniform magnetic field

I = current and A = Area

Diameter = 19cm = 0.19m so, radius = 0.19/2 = 0.095m

Area(A) = πr^(2) = πr^(2)

= π(0.095)^(2) = 0.0284 m^(2)

Now, B(earth)= 5x10^-5 T

While, we can ignore the angle because it's insignificant since the angle of the wire is oriented for maximum torque in the earth's field.

Now, if we arrange the formula to solve for charge (I):

I = (τ)/(A)(B)

I = (1.0x10^-3) / (0.0284)(5x10^-5)

I = 704A

6 0
3 years ago
2. Use the data table below to respond to the
forsale [732]

Answer:

car d

Explanation:

5555555555555555555

8 0
3 years ago
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