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zzz [600]
4 years ago
7

How much work would be required to lift a 12.0 kg mass up onto a table 1.15 meters high

Physics
1 answer:
astra-53 [7]4 years ago
8 0
Work = change of PE
PE = mgh 
Work = 12 kg * 10 * 1.15 m = 135 J 
You might be interested in
The "escape velocity" from Earth (the speed required to escape Earth's gravity) is 2.5 × 104 miles per hour. What is this speed
ANEK [815]

Answer:

2.5 × 10⁴ mi/h = 1.1 × 10⁴ m/s

Explanation:

Hi there!

We have the following equivalencies:

1 mile = 1609 m

1 hour = 3600 s

Then to convert miles to meters, we can multiply the given quantity in miles by ( 1609 m/ 1 mile) and we will obtain the same quantity in meters. In the same way, if we want to convert hours into seconds, we can multiply the given quantity in hours by (3600 s/ 1 hour) and we will obtain seconds.

Let´s convert miles per hour into m/s:

2.5*10^{4} \frac{mi}{h}  (\frac{1609m}{1mi})(\frac{1h}{3600s}) = 1.1 × 10⁴ m/s  (notice how the units mi and h cancel)

7 0
3 years ago
When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coeffic
Nuetrik [128]

Answer:

8.1\cdot 10^{-4} C^{-1}

Explanation:

The volumetric expansion of the liquid is given by

\Delta V=\alpha V_0 \Delta T

where

\alpha is the coefficient of volume expansion

V_0 is the initial volume

\Delta T is the change in temperature

For the liquid in this problem,

V_0 = 2.35 m^3\\\Delta T=48.5^{\circ}C\\\Delta V=0.0920 m^3

So we can solve the equation to find \alpha:

\alpha=\frac{\Delta V}{V_0 \Delta T}=\frac{(0.0920 m^3)}{(2.35 m^3)(48.5^{\circ}C)}=8.1\cdot 10^{-4} C^{-1}

7 0
4 years ago
A 13.0-V battery is connected in series with a switch, resistor and coil. If the circuit's time constant is 1.20 ´ 10-4 s and th
slavikrds [6]

Answer:

R = 5.20 Ω,    L = 6.24 10⁻⁴ H

Explanation:

The current in an RL circuit is

           I = \frac{E}{R} \ (1- e^{- t/ \tau } )

           τ = L / R

In the problem they indicate the value of the voltage, the current and the time constant, for which the resistance must be found

         

The stable current is when enough time has passed (t »τ) after closing the circuit, therefore the exponential term is very small and we can neglect it.

            I = E / R

            R = E / I

let's calculate

            R = 13.0 / 2.50

            R = 5.20 Ω

now with this value we can find the inductance of the coil

            τ = L / R

            L = τ  R

            L = 1.20 10⁻⁴   5.2

            L = 6.24 10⁻⁴ H

3 0
3 years ago
For the following statements, indicate whether the statement is true or false and explain in words your reasoning.
Nataly [62]

Answer:

a) If I increase the number of moving charged particles per cubic meter in a conductor, I will have to increase the cross sectional area of the conductor to get the same amount of current for the same electric field in the conductor.

FALSE

As we know that

I = neAv_d

so here if we increase the number of charge per unit volume then we need to decrease the area to get same amount of current through the wire.

b) The change in magnetic flux in a closed loop induces an EMF that opposes the change in magnetic flux.

TRUE

As per lenz law the direction of induced EMF is always opposite to the change in the flux due to which EMF is induced in the closed loop.

c) The magnetic force per unit length of two long, parallel, current-carrying conductors is repulsive if the currents are flowing in the opposite direction.

TRUE

When magnetic field of first wire will interact with other then it will exert force on it

so here the force will be on other wire such that the two wire will repel when current in the two wires is opposite in direction.

d) . The total magnetic flux through a closed surface is equal to μ0I enclosed

FALSE

magnetic flux is the number of field lines passing through a given area

So it is given as

B. A = 0

8 0
3 years ago
Which of the following is equal to the area under a velocity-time graph
Ksenya-84 [330]

-- The area under a velocity/time graph, between two points in time, is the difference in displacement during that period of time.

-- The area under a speed/time graph, between two points in time, is the distance covered during that period of time.

5 0
3 years ago
Read 2 more answers
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