400JA container with a mass of 5 kg is lifted to a height of 8 m. How much work is done
by the gravitational force?
O 50J
0400)
0-400J
o
zero
O -50)

Answer:

A: -10 kg•m/s

Explanation:

6•5= 30

-2•20=-40

-40+30= -10

-10 kg•m/s

Answer:

12) I = 800 kg m/ s ,  13) Δv = 31.25 m / s, 14)  F = -3600 N , 15)  Δv = 4.44 m/s

Explanation:

In this strange exercise you are asked to answer a series of questions

12) force F = 800N applied mass m = 80 kg has a speed change of DV = 10 m / s, calculate the momentum

           I = Δp

           I = m Δv

           I = 80 10

           I = 800 kg m / s

13) a mass of m = 0.80 kg, has an impulse of 25 N s, determines the change in speed

          I = m Δv

          Δv = I / m

          Δv = 25 / 0.8

          Δv = 31.25 m / s

14) a mass m + 1200 kg is carried from 25 m / s to 10 m / s in a period of 5.0 s, determine the force

          I = m ([tex]v_{f}[/tex] -v₀)

the proemd force for this impulse

          I = F t

we substitute

          F t = m (v_{f} -v₀)

          F = m (v_{f} -v₀ / t

          F = 1200 (10-25) / 5

          F = -3600 N

the negative sign indicates that the force is in the opposite direction of the movement

15) a mass = 90 kg with velocity v = 9.0 m has a momentum of I = 400N s, determines the change in velocity

           I = m Δv

           Δv = I / m

           Δv = 400/90

            Δv = 4.44 m / s

if they asked for the final speed it would be.

            I = m (v_{f} -v₀)

            v_{f} = v₀ + I / m

            v_{f} = 9 + 400/90

            v_{f} = 13.4 m / s

Explanation:

For balanced forces, the severity of the 2 factors is equal, while the magnitude of the 2 factors is unequal in the situation of unbalanced forces. The three separate forces operate in opposing ways in balanced forces. In unbalanced forces, on the other hand, independent forces either behave in the same or reverse direction.

Answer:

1.7x10^-10 N

Explanation:

F = G [(m_1)(m_2)]/(r^2)

F = force

G = Gravitational constant 6.67433x10^-11 (N*m^2)/kg^2

m_1 = mass first object

m_2 = mass second object

r = radius between the 2 objects

F = G[(2 kg*2 kg)/(1.25 m)^2]

F = 1.7x10^-10 N

Answer:

hjg567 A 567

Explanation:

An electric switch is a device that interrupts the electron flow in a circuit. Switches are primarily binary devices: either fully on or off and light switches have a simple design. When the switch is turned off, the circuit breaks and the power flow is interrupted. Circuits consist of a source of power and load.

Answer:

(1)Work done by snail is 382.5 N

(2)Work done by ELEPHANT is 25160 N

(3)Work done by HANS is 2420 N

(4)Work done on block is 433 N

Explanation:

The work done due to force F applied at an angle θ from the horizontal to the body is give by

Work done = Fscosθ where, s is distance traveled by body

Case 1: F= 255N , s= 3.00m and θ = 60.[tex]0^0[/tex]

Work done = Fscosθ = 255 x 3.00 x cos60.[tex]0^0[/tex] = 382.5N

thus work done by snail is 382.5 N

Case 2: F= 200N , s= 150m and θ = 33.[tex]0^0[/tex]

Work done = Fscosθ = 200 x 150 x cos33.[tex]0^0[/tex] = 25160N

thus work done by elephant is 25160 N

Case 3: F= 22.9N , s= 129m and θ = 35.[tex]0^0[/tex]

Work done = Fscosθ = 22.9 x 129 x cos35.[tex]0^0[/tex] = 2420N

thus work done by Hans is 2420 N

Case 4: F= 100N , s= 5m and θ = 30.[tex]0^0[/tex]

Work done = Fscosθ = 100 x 5.00 x cos30.[tex]0^0[/tex] = 433N

thus work done on block is 433 N

(1)  The work done by the sailor is 382.5 J.

(2)  The work done by elephant is 25160 J.

(3) The work done by Hans upon the backpack is 2420 J.

(4)  The required work done on block is 433 J.

Let us solve these questions in parts. All these questions are based on the work done. The work done due to force F applied at an angle θ from the horizontal to the body is give by,

[tex]W =F \times s \times cos \theta[/tex]

Here, s is the distance covered by the body.

(1)

Given data:

F= 255N , s= 3.00m and θ = 60.

The work done is calculated as,

W = Fscosθ

W= 255 x 3.00 x cos60 = 382.5 J.

Thus, the work done by the sailor is 382.5 J.

(2)

Given data:

F= 200N , s= 150m and θ = 33.

The work done by the elephant is calculated as,

W = Fscosθ

W= 200 x 150 x cos33. = 25160 J

Thus, the work done by elephant is 25160 J.

(3)

Given data:

F= 22.9N , s= 129m and θ = 35.

Then the work done upon the backpack is calculated as,

W= Fscosθ

W= 22.9 x 129 x cos35. = 2420 J

Thus, the work done by Hans upon the backpack is 2420 J.

(4)

Given data:

F= 100N , s= 5m and θ = 30.

The work done is calculated as,

W = Fscosθ

W= 100 x 5.00 x cos30. = 433 J

Thus, the required work done on block is 433 J.

Learn more about the work done here:

https://brainly.com/question/13662169

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

[tex]\dfrac{dF_{net}}{df}=0[/tex]

We need to calculate the unknown force

Using formula of net force

[tex]\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}[/tex]

Put the value into the formula

[tex]\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}[/tex]

[tex]\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}[/tex]

The magnitude of net force,

[tex]F_{net}=\sqrt{F_{x}^2+F_{y}^2}[/tex]

[tex]F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}[/tex]

[tex]F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}[/tex]

[tex]F_{net}=\sqrt{F^2+4899.78+36.232F}[/tex]

On differentiating w.r.to F

[tex](\dfrac{dF_{net}}{dF})^2=2F+36.232[/tex]

[tex]0=2F+36.232[/tex]

[tex]F=-\dfrac{36.232}{2}[/tex]

[tex]F=-18.116\ lb[/tex]

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.

Following are the solution to the given question:

Calculating the Net force on flange:

[tex]\overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 70 \cos 30^{\circ} -40 ) \hat{i}+(70 \sin 30^{\circ} - F \sin 45^{\circ} )\hat{y} \\\\ \overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 20.62 ) \hat{i}+(35 -F\sin 45^{\circ})\hat{y} \\\\ [/tex]

Calculating the magnitude:

[tex]= \sqrt{f_{X}^2 +f_{y}^{2}}\\\\ = \sqt{( F \cos 45^{\circ} + 20.62 )^2+(35 -F\sin 45^{\circ})^2} \\\\ = \sqt{( F^2+20.62^2+ 35^2+ 41.24 F \cos 45^{\circ} -70 F\sin 45^{\circ})}\\\\ = \sqt{( F^2+20.33F +1650.1844)} \\\\[/tex]

Differentiate the value

[tex]\to \frac{F_{net}}{df}=0 [/tex]

[tex]\to 2F-20.33=0\\\\ \to 2F=20.33\\\\ \to F=\frac{20.33}{2}\\\\ \to F= 10.165\ lb [/tex]

Learn more:

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Answer:

60n left

Explanation:

Do

Answer:

Momentum, p = 23250 kg m/s

Explanation:

Given that

Mass of a car, m = 1550 kg

Speed pf car, v = 15 m/s

We need to find the momentum of the car. The formula for the momentum of an object is given by :

p = mv

Substituting all the values in the above formula

p = 1550 kg × 15 m/s

p = 23250 kg m/s

So, the momentum of the car is 23250 kg m/s.

Answer:

[tex]from \: newton \: law \: of \: gravitation \\ F = \frac{GMm}{ {r}^{2} } \\ G = 6.67 \times {10}^{ - 11} \\ M = 0.5 \: kg \\ m = 0.33 \: kg \\ r = 0.002 \: m \\ substitute \\ F = \frac{(6.67 \times {10}^{ - 11}) \times (0.5) \times (0.33)) }{ {(0.002)}^{2} } \\ = 2.75 \times {10}^{ - 6} N[/tex]

Answer:

hope this will help

Explanation:

if your title doesn't have enough information, make a list of the key words ... If you are not sure what should be included in each summary sentence, use ... Often you can present the hypothesis and the supporting reasoning in one paragraph. ... Describe specific parts of the procedure or data that contributed to your learning.  

Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/191502-7how-do-the-data-support-your-claim-above-explain-your-statement.html

Answer:

 F_{net-x} = 12.32 N ,   F_{net-x} = -0.14 N  

Explanation:

For this case we apply Newton's second law on each axis,  see attachment

X axis

       F cos 40 - fr = m a

Y Axis  

      F sin 40 +N - W = 0

the net force is

X axis

     [tex]F_{net-x}[/tex] = F - fr

     F_{net-x} = 20 cos 40 - 3

     F_{net-x} = 12.32 N

Y Axis  

     F_{net-y} = F sin 40 + N - W

     F_{net-x} = 20 sin 40 + 7 - 20

     F_{net-x} = -0.14 N

we see that the force in this axis is approximately zero

Answer:

how the universe began

Explanation:

time never stops or starts, we only think it starts when we start recording it

Answer:

40 cm

hope it's help you but you can choose if your ans. is -40 cm

Answer:

??

Explanation:

not completely sure hold on

Answer:

-43J

Explanation:

Answer:

Center

Explanation:

The center is the tallest player on each team, playing near the basket. On offense, the center tries to score on close shots and rebound. But on defense, the center tries to block opponents' shots and rebound their misses.

Answer:

C

Explanation:

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

[tex]K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}[/tex]

Final kinetic energy is :

[tex]K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}[/tex]

Now, fraction of initial kinetic energy loss is :

[tex]Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25[/tex]

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

Explanation:

The compass needle moved when the wire was connected to the battery. The important point here is that the needle is affected by the wire only when both ends of the wire are connected to the battery because only at this time is current flowing through the circuit.

Answer:

this answer you question

Answer:

Explanation:

A(n) generator changes mechanical energy into electrical. A(n) motor changes electric energy into mechanical. A(n) transformer changes the voltage of an alternating current. A(n) step-down transformer has more loops in the primary coil than in the secondary coil. A(n) step-up transformer has fewer loops in the primary coil than in the secondary coil.​

Hope it's helpful!

Answer:

u will need 1/2 cells

Explanation:

Answer:

c

Explanation:

Answer:

a = 5 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

[tex]v_{f}=v_{o}+a*t[/tex]

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 10 [m/s]

t = time = 2 [s]

a = acceleration [m/s²]

Now replacing:

[tex]20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ][/tex]

Answer:

a, 22 kg and 22 kg

b, 215.8 N and 440 N

Explanation:

a

The mass of the ball remains constant and unchanged irrespective of where it has been to, need to go or is going. So, basically the mass of the ball on earth is as the same mass of the ball on the said planet, 22 kg

b

The weight of any object factors in the acceleration due to gravity of the said area(or planet).

W = mg, with m being the mass and g being the acceleration due to gravity.

On earth

W = 22 * 9.81 = 215.8 N

On the said planet,

W = 22 * 20 = 440 N

Do therefore, the weight is 215.8 N on earth and 440 N on the planet

Answer:

5m

Explanation:

khan academy

Vertical displacement of the circus performer is 5 m.

Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.

A circus performer walked up and to the right for a total displacement of 10 m, start to end along a diagonal tightrope angled 30° degree above the ground.

Vertical displacement = 10 sin 30° = 5 m

Vertical displacement of the circus performer is 5 m.

To learn more about displacement refer to the link:

brainly.com/question/11934397

#SPJ2

Answer:

I believe the answer is specific gravity

Explanation:Hope this helps :)

Answer:

Yayyyyyy thank you ≥﹏≤.!!!!!!!!!!!!!

Answer:

mesosystem

Explanation:

The mesosystem involves relations between microsystems or connections between contexts.

Answer:

Use equation 1/2 * m * v²

Explanation:

where m = mass in kg

and v = velocity in m/s

plug accordingly.