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olchik [2.2K]
3 years ago
9

Lee pushes horizontal with a force of 79 N on a 34Kg mass for 10 m across a floor. Calculate the amount of work lee did

Physics
1 answer:
Furkat [3]3 years ago
3 0
First you have to know the formula for work which is force times (x) distance which is equal to Newtons times (x) mass answer is in joules which is tge unit
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An elevator is moving is an upwards
Minchanka [31]

Answer:

The elevator's free-body diagram has three forces, the force of gravity, a downward normal force from you, and an upward force from the tension in the cable holding the elevator. The combined system of you + elevator has two forces, a combined force of gravity and the tension in the cable.

Explanation:

6 0
3 years ago
Disorder in the universe increases because
Veronika [31]
I think the answer would be c
7 0
3 years ago
Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,
REY [17]

Answer:

a) 238U, 40K and 87Rb, b)   235U and to a lesser extent 40K , c)  he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0
3 years ago
What material would you use to build countertops
babunello [35]

Answer:   Natural stone

Explanation:

6 0
2 years ago
Read 2 more answers
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
3 years ago
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