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jasenka [17]
3 years ago
10

Name six kinds of simple machines.

Physics
2 answers:
Gnom [1K]3 years ago
8 0

Answer:

the wheel and axle, the lever, the inclined plane, the pulley, the screw, and the wedge

Explanation:

Nutka1998 [239]3 years ago
3 0
The wheel, Axel, Lever, Inclined Pane, pulley, screw, and wedge!
You might be interested in
Here's a question from ~ [ AIEEE 2002 ]
lyudmila [28]
  • r=150m
  • coefficient of friction=\mu=0.6

As car is avoid skidding

\\ \sf\hookrightarrow \dfrac{mv^2}{r}=\mu mg

  • Cancel m

\\ \sf\hookrightarrow \dfrac{v^2}{r}=\mu g

\\ \sf\hookrightarrow v^2=\mu rg

\\ \sf\hookrightarrow v^2=0.6(10)(150)

\\ \sf\hookrightarrow v^2=60(150)

\\ \sf\hookrightarrow v^2=900

\\ \sf\hookrightarrow v=30ms^{-1}

Done

7 0
2 years ago
Read 2 more answers
A spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated
I am Lyosha [343]

Answer:

A) 1.4167 × 10^(-11) F

B) r_a = 0.031 m

C) E = 3.181 × 10⁴ N/C

Explanation:

We are given;

Charge;Q = 3.40 nC = 3.4 × 10^(-9) C

Potential difference;V = 240 V

Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m

A) The formula for capacitance is given by;

C = Q/V

C = (3.4 × 10^(-9))/240

C = 1.4167 × 10^(-11) F

B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.

C = (4πε_o)/(1/r_a - 1/r_b)

Rearranging, we have;

(1/r_a - 1/r_b) = (4πε_o)/C

ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m

Plugging in the relevant values, we have;

(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))

(1/r_a) - 24.3902 = 7.8501

1/r_a = 7.8501 + 24.3902

1/r_a = 32.2403

r_a = 1/32.2403

r_a = 0.031 m

C) Formula for Electric field just outside the surface of the inner sphere is given by;

E = kQ/r_a²

Where k is a constant value of 8.99 × 10^(9) Nm²/C²

Thus;

E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²

E = 3.181 × 10⁴ N/C

3 0
3 years ago
Most geologists believe that the dinosaurs became extinct 65 million years ago when a large comet or asteroid struck the earth,
Alex17521 [72]

Answer:

A) 1.67 x 10 ⁻⁶ m/s

B)5.59 x 10^-^9 %

Explanation:

A)

Given:

d = 5.0 km,

mₐ = 2.5 x 10^1^4 kg

u₁ = 4.0 x 10⁴ m/s

m_n = 5.98 x 10 ²⁴ kg

Solve using kinetic conserved energy

mₐ x u₁ + m_n  x u₂ = uₓ x (mₐ + m_n )

(2.5 x 10^1^4) (4.0 x 10⁴ )+ (5.98 x 10 ²⁴ )(0) = uₓ x (2.5 x 10^1^4 + 5.98 x 10 ²⁴ )

uₓ = ( 2.5 x 10^1^4 x 4.0 x 10⁴ ) / (2.5 x 10^1^4 + 5.98 x 10 ²⁴ )    

uₓ = 1.67 x 10 ⁻⁶ m/s

B) Assuming earth radius as a R = 1.5 x 10 ¹¹ m

t = 365 days  x 24 hr / 1 day x 60 minute / 1 hr x 60s / 1 minute = 31536000 s

t = 31536000 s

D = 2 π R = 2 π( 1.5 x 10 ¹¹ )

D = 9.4247 x 10 ¹¹ m

u₂  = D / t  = 9.4247 x 10 ¹¹  / 31536000

u₂ =  29885.775 m/s

% = (  1.67 x 10 ⁻⁶ m/s ) / (29885.775 m/s) x 100

% = 5.59 x 10^-^9 %

5 0
3 years ago
What is the magnitude of the electric field at the origin produced by a semi-circular arc of charge = 7.6 μc, twice the charge o
Anna [14]

Orient the semi-circle arc such that it is symmetric with respect to the y-axis. Now, by symmetry, the electric field in the x-direction cancels to zero. So the only thing of interest is the electric field in the y-direction.  


dEy=kp/r^2*sin(a) where k is coulombs constant p is the charge density r is the radius of the arc and a is the angular position of each point on the arc (ranging from 0 to pi. Integrating this renders 2kq/(pi*r^3). Where k is 9*10^9, q is 9.8 uC r is .093 m

I answeared your question can you answear my question pleas

6 0
3 years ago
An electron is projected with an initial speed Vo = 5.35x10^6 m/s into the uniform field between the parallel plates. The direct
mart [117]

Answer:

 E = 3.04 10⁻⁵ N / C

Explanation:

In this problem we can use the kinematics to find how long it takes the electron to travel the plates

Let's start by reducing the magnitudes to the SI system

              vₓ = 5.35 10⁶ m / s

              x = 2 cm = 2 10⁻² m

              y = 1 cm = 1 10⁻² m

             x = vₓ t

             t = x / vₓ

             t = 2 10⁻² / 5.35 10⁶

             t = 3,738 10⁻⁹ s

This time is also the time it takes for vertical movement to go from the center to the plate, let's look for acceleration with Newton's second law

              F = m a

              a = F / m = e E / m

              y = v_{oy} + ½ a t²

              v_{oy} = 0

We replace

              y = ½ e / m E t²

              E = 2 y m / e t²

Let's calculate

            E = 2  1 10⁻²  9.1 10⁻³¹ / (1.6 10⁻¹⁹  3,738 10⁻⁹)

            E = 18.2 10⁻³³ / 5.98 10⁻²⁸

            E = 3.04 10⁻⁵ N / C

5 0
3 years ago
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