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2 years ago

If a car is rounding a flat curve on a highway, what is the centripetal force on the car?

Physics

1 answer:

tigry1 [53]2 years ago

8 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.

For a level curve, the centripetal force will be supplied by the friction force between the tires and roadway.

A banked curve can supply the centripetal force by the normal force and the weight without relying on friction.

Explanation:

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A Heavy crate applied a force of 1500 N on a 25m2 piston. What force needs to be applied on a 0.8m2 piston to lift the crate?

Aleks04 [339]

25/1500 is equal to 0.8/x
0.8*1500 is equal to 1200
1200/25 is equal to 48 N

7 0

3 years ago

Read 2 more answers

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

7 0

2 years ago

A hayride wagon is going down a spooky country road at 15 m/s when a Scarecrow appears in the roadway. The man at the wheel of t

nexus9112 [7]

Answer:

d = 68.18 m

Explanation:

Given that,

Initial velocity, u = 15 m/s

Finally it comes to stop, v = 0

Acceleration, a = -1.65 m/s²

Time, t = 2.5 s

We need to find the distance covered by the hayride before coming to a stop. Let d is the distance covered. Using third equation of motion to find it :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(15)^2}{2\times -1.65}\\\\d=68.18\ m$

So, the hayride will cover a distance of 68.18 m.

6 0

3 years ago

Calculate the density of a tin of mass 100g whose dimensions are 2cmx5cmx

Luda [366]

Answer:

your question in not complete.

you need to the high too.

7 0

2 years ago

If a mass of .343 kg moves down a 5m ramp in 5.8 seconds, what is the velocity and KE developed by the moving mass

Soloha48 [4]

Velocity 0.86m/s

0.13J

Explanation:

Given parameters:

Mass = 0.343kg

distance = 5m

time taken = 5.8s

Unknown:

Velocity of mass = ?

Kinetic energy = ?

Solution:

Velocity is the rate of change of displacement with time. It is a vector quantity that shows magnitude and direction.

Mathematically;

Velocity = $\frac{displacement}{time}$

Velocity = $\frac{5}{5.8 }$ = 0.86m/s

Kinetic energy is the energy due to the motion of a body. It is expressed mathematically as:

Kinetic energy = $\frac{1}{2} m v^{2}$

m is the mass

v is the velocity

Kinetic energy = $\frac{1}{2 } x 0.343 x 0.86^{2}$ = 0.13J

learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

7 0

3 years ago