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2 years ago

The sprinter experiences ___________ from A to B to C.

Physics

2 answers:

Trava [24]2 years ago

5 0

motion from A to B to C.....

Nitella [24]2 years ago

4 0

The sprinter experiences distance from A to B to C

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A steel cable with Cross Sectional Area 3.00cm² has an elastic limit of 2.40 x 10^8pascals. Find the maximum upward acceleration

bazaltina [42]

Answer:

Stress = F / A force per unit area

A = 3.00 cm^2 = 3 E-4 m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N max force applied

F/3 = 2.4E4 N if force not to exceed limit (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2 about 2 g

3 0

2 years ago

A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capp

Bond [772]

Answer:

= 925.92 N

≅ 926N

Explanation:

Pressure due to car = pressure due to applied force

12000/18^2 = Force / 5^2

force = 12000 * 25/ 324

= 925.92 N

For equilibrium

Pressure1 = Pressure2

A1F1 = A2F2

12000*pi*(5^2) = F2 ( pi)*(18^2)

so, F2 = Applied force to lift car = 925.92 N

Pascal's principle

Pressure1 = Pressure2

F1/A1 = F2/A2 (F=force and A=area)

A1 =Pi*(0.05)²

A2 =Pi(0.18)²

F2=12000

F1 = 12000*(0.05)² / (0.18)² = 926N

7 0

2 years ago

A carrot hangs from the ceiling by a rope,

valentina_108 [34]

Answer: Diagram B

Explanation:

A free body diagram shows the forces acting on an object in a certain scenario.

In this scenario there are two forces acting on the carrot: the Tension force (Ft) from the rope that the carrot is hanging from and Gravitational force(Fg) which is pulling the carrot to the Earth.

The diagram depicting this is diagram B.

7 0

2 years ago

Read 2 more answers

19. A person pushes with 6.0 N for 4.0 seconds on a 2.0 kg object.

hram777 [196]

Answer:24NS

Explanation:

Impulse=force x time

Impulse=6 x 4

Impulse=24NS

5 0

3 years ago

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

2 years ago