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sergejj [24]
1 year ago
11

The free-body diagram below represents the force of several vehicles driving across a bridge. Assume that the bridge is in stati

c equilibrium and that it has zero weight, and solve for the unknown reaction force.
Physics
1 answer:
krek1111 [17]1 year ago
7 0

In order to calculate the unknown reaction force, we need to know that the sum of forces pointing down is equal the sum of the forces pointing up, so all forces will be in equilibrium.

So we have:

\begin{gathered} 17800+150000+13200=121000+x \\ 181000=121000+x \\ x=181000-121000 \\ x=60000 \end{gathered}

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A rocket for use in deep space is to have the capability of boosting a total load (payload plus the rocket frame and engine) of
choli [55]

Answer:

<em>13.54 tons</em>

Explanation:

Let f be the amount of fuel oxidizer needed

v be the speed

The relationship between them is inverse in nature i.e

f ∝ 1/v

f = k/v

If a rocket for use in deep space is to have the capability of boosting a total load (payload plus the rocket frame and engine) of 3.25 metric tons to a speed of 10,000 m/s, then f = 3.25 when v  = 10,000

Substitute and get k

k = fv

k = 3.25 * 10,000

k = 32500

To get the amount of fuel oxidizer required to produce a speed of 2400m/s, we will find f when v = 2400m/s

Recall that f = k/v

f = 32500/2400

f = 13.54 metric tons

<em>Hence the fuel plus oxidizer that will be required is 13.54 tons</em>

4 0
3 years ago
G davisson and germer performed their experiment with a nickel target for several electron bombarding energies. at what angles w
Anton [14]
If the object, ends up with a positive charge, then it is missing electrons. if it is missing electrons, then it must have been removed form the object during the rubbing process.
5 0
3 years ago
Each proton-proton cycle generates 26.7 MeV of energy. If 9.9 Watts are generated via by the proton-proton cycle, how many billi
lina2011 [118]

Answer:

4.635 *10^12 Neutrinos

Explanation:

Here in this question, we are to determine the number of neutrinos in billions produced, given the power generated by the proton-proton cycle.

We proceed as follows;

In proton-proton cycle generates 26.7 MeV of energy and in this cycle two neutrinos are produced.

From the question, we are given that

Power P = 9.9 watts = 9.9 J/s

Watts is same as J/s

The number of proton-proton cycles required to generate E energy is N = E / E '

Where E ' = Energy generated in proton-proton cycle which is given as 26.7 Mev in the question

Converting Mev to J, we have

= 26.7 x1.6 x10 -13 J

To get the number N which is the number of proton-proton cycle required, we have;

N = 9.9 /(26.7 x1.6 x10^-13) = 2.32 * 10^12

Since we have two proton cycles( proton-proton), it automatically means 2 neutrinos will be produced.

Therefore number of neutrions produced = 2 x Number of proton-proton cycles = 2 * 2.32 * 10^12 = 4.635 * 10^12 neutrinos

8 0
3 years ago
What does biological pest control involve?
bogdanovich [222]
Biological control<span> is a </span>bioeffector<span>-method of </span>controlling pests<span> such as </span>insects<span>, </span>mites<span>, </span>weeds<span> and </span>plant diseases, using other living organisms.
<span>Biological pest control relies on</span> natural mechanisms including predation, parasitism, or herbivory<span>, </span>but typically also involves an active human management role.
5 0
3 years ago
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
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