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STALIN [3.7K]
2 years ago
10

The standard acceleration (at sea level and 45◦ latitude) due to gravity is 9.806 65 m/s2. What is the force needed to hold a ma

ss of 2 kg at rest in this gravitational field? How much mass can a force of 1 N support?
Physics
1 answer:
erma4kov [3.2K]2 years ago
5 0

a) Remember Newton's second Law:

F = m*a

This means that the forces that is being exerted over an object is equal to the mass of the object times the acceleration that it has.

In this case, in order to hold the mass you need a force with the same magnitude but opposite direction to the gravitational force. The magnitude of this force would be the mass of the object times the gravitactional  acceleration:

F = 2kg*9.80665m/s² = 19.6133 N

b) Using again Newton's second Law, we can issolate mass from that equation:

m = F/a

Then, the mass that a force of 1N can support is equal to:

m = 1N/9.80665m/s² = 0.102 kg

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If you found yourself on the see-through side of this one-way mirror what is the best way you could prevent someone on the other
Taya2010 [7]

Answer:

If there is any sheets or padded material in this room you can cover the window, you could turn off all the lights if there is a light switch in the room,   you could try to bring a bright flashlight in and shine it into the other room(try to annoy the person watching you so they leave), act really boring and hopefully make the other person lose interest.

Explanation:

(hint) If you actually get in a situation like this place your fingernail against the mirror or glass you think could possibly be a one-way mirror. If there's a gap between your nail and the mirror, it's most likely a genuine mirror :)

7 0
1 year ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

J=\frac{E}{\rho}      (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

J=\frac{I}{A}        (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}

Next, you replace for all variables:

I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A

hence, the current in the wire is 4.75A

4 0
2 years ago
An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t
Setler [38]

Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

initial angular velocity is termed as positive angular velocity

\omega = positive

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

\alpha = negative

so here correct answer is

B. W is positive and a is negative

8 0
2 years ago
Round your answers to one decimal place. This parallel circuit has two resistors at 15 and 40 ohms. What is the total resistance
goldenfox [79]
-- With two resistors in parallel, the total effective resistance is
the reciprocal of  (1/R₁ + 1/R₂).

1/R₁ + 1/R₂  = 1/15 + 1/40

= 8/120 + 3/120

= 11/120

So the total effective resistance is 120/11 = 10.9 ohms .

Current = (voltage) / (resistance)

= 12 / (120/11)

= (12 · 11) / 120

= 132/120  =  1.1 Amperes  
4 0
2 years ago
Read 2 more answers
A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re
Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Pressure at reservoir = 10 atm

T_1 = Temperature at reservoir = 300 K

P_2 = Pressure at exit = 1 atm

T_2 = Temperature at exit

R_s = Mass-specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

For isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

The density of the flow at the exit is 2.2721 kg/m³

4 0
2 years ago
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