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2 years ago

Where is the centripetal force directed on a banked curve?

Physics

2 answers:

Alex777 [14]2 years ago

7 0

Answer:

The net force on a car traveling around a curve is the centripetal force, Fc = m v2 / r, directed toward the center of the curve.

Explanation:LET ME KNOW IF ITS WRONG. HAVE A NICE DAY!!!!!!!!!!

san4es73 [151]2 years ago

6 0

Answer: The net force on a car traveling around a curve is the centripetal force, Fc = m v2 / r, directed toward the center of the curve. The centripetal force is directed at right angles to the motion and also along the radius towards the centre of the circular path. In order for a car to move around a circular track, something must provide a centipetal force. One source of the centripetal force is friction between the tires and the road. This means that the tires do not need to supply any friction at all to keep the car from slipping sideways. ...

Hope this helps... stay safe and have a great day!!! :D

You might be interested in

Calculate the molarity of a 10. 0% cacl₂ solution. The density of the solution is 1. 0835 g/cm³.

brilliants [131]

The molarity of 10% CaCl2 is 0.9%

concentration of the given salt CaCl₂ = 10%

Density of a solution = 1.0835 g/cm³

Volume = m / d

= 100 / 1.0835

= 92.29 litres

Density = mass / volume

1.0835 × 92.29 = mass

mass = 99.99 gram

Thus the molarity can be calculated by = moles of solute / volume of solution multiplied by 100

= 0.9008/ 92.29 X 100 %

= 0.009 X 100 %

= 0.9 %

The molarity of 10% CaCl2 is 0.9%

To know more about density and molarity you may visit the link which is mentioned below:

brainly.com/question/10710093

#SPJ4

5 0

1 year ago

Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

5 0

3 years ago

The current in a series circuit is 19.3 A. When an additional 7.40-Ω resistor is inserted in series, the current drops to 13.4 A

Bogdan [553]

Answer:

16.8ohms

Explanation:

According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.

Mathematically, V = IRt where;

V is the voltage across the circuit

I is the current

R is the effective resistance

For a series connected circuit, same current but different voltage flows through the resistors.

If the initial current in a circuit is 19.3A,

V = 19.3R... (1)

When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;

V = 13.4(7.4+R)... (2)

Note that the initial resistance is added to the additional resistance because they are connected in series.

Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;

19.3R = 13.4(7.4+R)

19.3R = 99.16+13.4R

19.3R-13.4R = 99.16

5.9R = 99.16

R= 99.16/5.9

R = 16.8ohms

The resistance in the original circuit will be 16.8ohms

5 0

3 years ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>

Let's draw the free body diagram of the system using the given data.
From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

$N_x=86.62N$

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

$N_y=F_V=mg-Tsin59\\$