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2 years ago

13

. A horizontal steel spring has a spring constant of 40.0 N/m. What force must be applied to the spring in order to compress it

by 10.0 cm?

1 answer:

ANEK [815]2 years ago

7 0

Ans 4 more to be exact

Explanation:

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Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the

Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0

2 years ago

Investigators are exploring ways to treat milk for longer shelf life by using pulsed electric fields to destroy bacterial contam

Georgia [21]

Answer:

C = 3.77*10⁻¹⁰ F = 377 pF

Q = 1.13*10⁻⁵ C

Explanation:

Given

D = 8.0 cm = 0.08 m

d = 0.95 cm = 0.95*10⁻² m

k = 80.4 (dielectric constant of the milk)

V = 30000 V

C = ?

Q = ?

We can get the capacitance of the system applying the formula

C = k*ε₀*A / d

where

ε₀ = 8.854*10⁻¹² F/m

and A = π*D²/4 = π*(0.08 m)²/4

⇒ A = 0.00502655 m²

then

C = (80.4)*(8.854*10⁻¹² F/m)*(0.00502655 m²) / (0.95*10⁻² m)

⇒ C = 3.77*10⁻¹⁰ F = 377 pF

Now, we use the following equation in order to obtain the charge on each plate when they are fully charged

Q = C*V

⇒ Q = (3.77*10⁻¹⁰ F)*(30000 V)

⇒ Q = 1.13*10⁻⁵ C

7 0

2 years ago

A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s

Anastaziya [24]

Answer:

the correct answer is c v₁> 12.5 m / s

Explanation:

This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.

v² = v₀² + 2 a₁ x

as part of rest v₀ = 0

a₁ = v² / 2x

a₁ = 25² / (2 120)

a₁ = 2.6 m / s²

now we can find the velocity for the distance x₂ = 60 m

v₁² = 0 + 2 a1 x₂

v₁ = Ra (2 2,6 60)

v₁ = 17.7 m / s

these the speed at 60 m

we see that the correct answer is c v₁> 12.5 m / s

5 0

2 years ago

The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al

SashulF [63]

Answer:

7.9 $\frac{m}{s^{2} }$

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

$\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\$

Using algebra, we can rearrange our equation to find the final acceleration, $a_{2}$ :

$a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\$

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 $\frac{m }{s^{2} }$ = 8.33 $\frac{m }{s^{2} }$

Plug everything in:

7.9 $\frac{m }{s^{2} }$ = $\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}$

(1590kg the initial weight plus the weight of the added passenger)

8 0

2 years ago

Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The mass o

insens350 [35]

Answer:

295.42 N

Explanation:

From Newton's law of universal gravitation.

F = Gmm'/r².................. Equation 1

Where F = Gravitational force, G = Universal constant, m = mass of the human, m' = mass of mass, r = radius of mass.

Given: m = 80 kg, m' = 6.4×10²³ kg, r = 3.4×10⁶ m.

Constant: G = 6.67×10⁻¹¹ Nm²/Kg²

Substitute into equation 1

F = 6.67×10⁻¹¹(80)(6.4×10²³ )/( 3.4×10⁶)²

F = 3415.04×10¹²/(11.56×10¹²)

F = 3415.04/11.56

F = 295.42 N

Hence the gravitational force = 295.42 N

5 0

2 years ago