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Alex
3 years ago
7

A man can use a snowblower to move snow in 5 minutes. If he moves it with a shovel, will take 20 minutes. He will use more power

with which process?
Physics
1 answer:
Neko [114]3 years ago
6 0

Answer:

the second process of using the the shovel to move the snow as it takes a longer time meaning more power is expended

You might be interested in
A meter stick is free to rotate about an axis through one of its end. Find the force F needed to balance this meter stick if the
Mumz [18]

Answer:

Explanation:

Component of force perpendicular to stick

= F Sin 60°

=√3 / 2 F.

Taking torque about the other end

= √3 / 2 F x 1 Nm

Weight of stick = 60 gm

= 60 x 10⁻³ kg

= 60 x 10⁻³ x 9.8 N

= .588 N

This weight will act from the middle point of stick so torque about the

other end

= .588 x 1 Nm

Balancing these two torques we have

.588 = √3 /2 F

F=\frac{2\times0.588}{\sqrt{3} }

F = 0.679 N

6 0
3 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
2 years ago
A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b
jonny [76]

Answer:

Part a)

I = 3.16 \times 10^{-5} W/m^2

Part b)

P_o = 0.162 Pa

Explanation:

Part a)

Level of sound = 75 dB

now we know that

L = 10 Log\frac{I}{I_0}

here we know that

I_0 = 10^{-12} W/m^2

now we have

75 = 10 Log(\frac{I}{10^{-12}})

I = 3.16 \times 10^{-5} W/m^2

Part b)

Intensity of sound wave is given as

I = \frac{1}{2}\rho A^2\omega^2 c

here we know that

A = \frac{P_o}{Bk}

so we have

I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c

I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}

now we know

\rho = 1.2 kg/m^3

c = 340 m/s

B = 1.4 \times 10^5 Pa

now we have

3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}

P_o = 0.162 Pa

5 0
3 years ago
A collapsible plastic bag (figure below) contains a glucose solution. If the average gauge pressure in the vein is 1.29 A collap
expeople1 [14]

Answer:

0.125 m

Explanation:

Pressure in fluids is given as the product of density, height and acceleration due to gravity and expressed as

P=hdg

Where h is the height, d is density, g is acceleration due to gravity and P is pressure.

Making h the subject of formula then

h=P/dg

Given specific gravity of a substance, its density is equal to specific gravity multiplied by density of water. Taking density of pure water as 1000 kg/m³ then the density of reference fluid will be 1.05*1000=1050 kg/m³

Substituting pressure with 1.29*10³ pa as given then taking g as 9.81 m/s² then

H=1.29*10³÷(9.81*1050)=0.1252366389981068880151448958788408329692m

Rounded off, the height is approximately 0.125 m

8 0
3 years ago
A circular coil of five turns and a diameter of 30.0 cm is oriented in a vertical plane with its axis perpendicular to the horiz
____ [38]

The horizontal component of the magnetic field is 12.6 μT.

The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.

A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.

Let B be the magnetic field and R be the radius of the circular coil.

Then the horizontal component of the Earth's magnetic field is given as:

B(h) = B(coil) = μ₀ NI / 2R

B(h) = (4π × 10⁻⁷ ) (5)(0.6) / 0.3

B(h) = 12.6 μT

Learn more about magnetic field here:

brainly.com/question/14411049

#SPJ4

5 0
1 year ago
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