An object is released from a rocket moving upwards at a speed of 8.9 m / s.​
What is the downward acceleration of this object?

Answers

Answer 1

Answer:

9.81 m/s²

Explanation:

It is given that the upward speed of the object is 8.9 m/s.

An object is released.

Now, we know that if the object is released then it will move under the control of gravity which means that the initial velocity of the rocket will not affect the acceleration of the object.

Hence, the downward acceleration of the object will be equal to g i.e. 9.81 m/s².


Related Questions

What is the net force?
60N left
4ON, right
60N, right
0N

Answers

Answer:

60n left

Explanation:

Do

Juans mother drives 7.25 miles southwest to her favorite shopping mall. What is the average velocity of her automobile if she arrives at the mark in 20min?

Answers

21.75 Miles Per Hour

I got this by multiplying 7.25(3) because I know 20 minutes is 1/3 of 1 he

F eeeeeeeereeeeeee points

Answers

Answer:

Yayyyyyy thank you ≥﹏≤.!!!!!!!!!!!!!

What is the force of gravitational attraction between an object with a mass of 2kg and another object that has a mass of 2kg and a distance between them of 2.5m

Answers

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

What does the Big Bang theory explain?
A. How time began
B. How the universe began
C. Both A and B
D. Neither A nor B

Answers

Answer:

how the universe began

Explanation:

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

Two similarities between balanced and unbalanced forces.

Answers

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.

What height would a 4 kg book need to be to have a potential energy of
235.2 J on earth?*
need help!!

Answers

Answer:

5.99 m  = 6 m

Explanation:

PE = m*g*h

235.2 J = (4 kg)(9.81 m/s^2)(h)

h = (235.2 J)/(9.81*4)

h = 5.99 m

h = 6 m

1. A sailor pulls a boat along a dock using a rope at an angle of 60.0º with the horizontal. How much work does the sailor do if he exerts a force of 255 N on the rope and pulls the boat 3.00 m?

W = FdcosO =255 x 3 x cos 60 =


2. An elephant pushes with 2000 N at an angle of 33o above the horizontal on a load of trees. It then pushes these trees for 150 m. How much work did the elephant do?


3. Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35o above the horizontal to drag his backpack a horizontal distance of 129 m. Determine the work done upon the backpack.


4. If 100 N force has 30o angle pulling on a 15 kg block for 5 m. What’s the work?

Answers

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

The correct answer is 40 or - 40??

Answers

Answer:

40 cm

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

Calculate the difference in blood pressure between the feet and top of the head for a person who is 1.70 m tall.

Answers

Answer:

[tex]P_2 - P_1 = 1.8 * 10^4\ Pa[/tex]

Explanation:

Given

[tex]Height (h) = 1.70m[/tex]

Required

Determine the difference in the blood pressure from feet to top

This is calculated using Pascal's second law.

The second law is represented as:

[tex]P_2 = P_1 + pgd[/tex]

Subtract P1 from both sides

[tex]P_2 - P_1 = pgd[/tex]

Where

[tex]p = blood\ density = 1.06 * 10^3kg/m^3[/tex]

[tex]g = acceleration\ of\ gravity = 9.8N/kg[/tex]

[tex]d =height = 1.70m[/tex]

P2 - P1 = Blood Pressure Difference

So, the expression becomes:

[tex]P_2 - P_1 = 1.06 * 10^3 * 9.8 * 1.70[/tex]

[tex]P_2 - P_1 = 17659.6Pa[/tex]

[tex]P_2 - P_1 = 1.8 * 10^4\ Pa[/tex]

Hence, the difference in blood pressure is approximately [tex]1.8 * 10^4\ Pa[/tex]


1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?

Answers

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.

A 20N cart is being pulled to the right with an applied force of 20N at an angle of 40°. The normal force on the cart is 7N and the friction is 3N. Draw the free body diagram. What is the net force in the x and y direction of the cart?

Answers

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

When the torques on the wheel and axle equal zero, the machine has
_____

Answers

Answer:

8

Explanation:

Answer:

rotational equilibrium

Explanation:

Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acting on the flange is a minimum.

Answers

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:

brainly.com/question/15323099

Which waves are electromagnetic and can travel through a vacuum?

Light and heat waves
Longitudinal and transverse waves
Sound waves
Surface waves

Answers

Light and heat waves

I NEED HELP WITH 8 I WILL GIVE U BRAINLIEST

Answers

Answer:

u will need 1/2 cells

Explanation:

At the surface of a certain planet, the gravitational acceleration g has a magnitude of 20.0 m/s^2. A 22.0-kg brass ball is transported to this planet.
What is (a) the mass of the brass ball on the Earth and on the planet, and (b) the weight of the brass ball on the Earth and on the planet?

Answers

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

The x-axis of a position-time graph represents

Answers

Answer:

The y-axis represents position relative to the starting point, and the x-axis represents time.

Explanation:

pls help, i have no idea where to start

Answers

Answer:

Use equation 1/2 * m * v²

Explanation:

where m = mass in kg

and v = velocity in m/s

plug accordingly.

Explain why crinoid fossils can be found on the shore of Lake Michigan, which is a fresh water lake.

Answers

Answer:

Explanation:

Crinoid fossils can be found on the shore of Lake Michigan because during the Carboniferous period, all of what is now the United States, except for a small part of the upper Midwest, and all of the states along the East Coast, except for Florida, was covered by a warm, shallow inland sea(seawater), and since crinoid fossils live in sea water, they washed up on many places like Lake Michigan.

Answer:

Explanation:

Crinoid fossils can be found on the shore of Lake Michigan because during the Carboniferous period, all of what is now the United States, except for a small part of the upper Midwest, and all of the states along the East Coast, except for Florida, was covered by a warm, shallow inland sea(seawater), and since crinoid fossils live in sea water, they washed up on many places like Lake Michigan.

What are the characteristics of supernovae?
release of matter and energy
provide new material for future solar systems
move on to form a black hole or neutron star
might be seen from earth

Answers

Answer:

c

Explanation:

Can yall help me please

Answers

Answer:

physical

Explanation:

it is a real life model that was built

The density of a solid or liquid material divided by the density of water is called

Answers

Answer:

I believe the answer is specific gravity

Explanation:Hope this helps :)

Help me
Shown is a picture of two toy trains getting ready to collide. If the collision is non-elastic, what is the total momentum of

the train cars after the collision?

a

-10 kg·m/s


b

10 kg·m/s


c

70 kg·m/s


d

50 kg·m/s

Answers

Answer:

A: -10 kg•m/s

Explanation:

6•5= 30

-2•20=-40

-40+30= -10

-10 kg•m/s

Which players are usually the tallest on their team, and stay close to the basket so they can shoot and rebound the ball?
Mid-fielder
Center
Forward
Guard

Answers

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:

Numerical Problems
A bus is moving with the initial velocity 10 m/s. After 2 seconds, the velocity
becomes 20 m/s. Find the acceleration and distance moved by the bus.​

Answers

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]

A north magnetic pole brought near a south magnetic pole will...


A)
cancel the south pole.

B)
attract the south pole.

C)
repel the south pole.

Answers

It’s B) I hope this helped

what is the force of gravity attraction between an object with a mass of 0.5 kg and another object has a mass of 0.33 kg and a distance between Dam of 0.002 M ​

Answers

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]

Fill in the blanks in the following paragraph to correctly identify the properties of devices that use electromagnetism. (1 point)
A(n) ____________________ changes mechanical energy into ____________________. A(n) ____________________ changes electric energy into ____________________. A(n) ____________________ changes the voltage of an alternating current. A(n) ____________________ transformer has more loops in the primary coil than in the secondary coil. A(n) ____________________ transformer has ____________________ loops in the primary coil than in the secondary coil.​

Answers

Answer:

this answer you question

Answer:

generatorelectricalmotormechanicaltransformerstep-downstep-upfewer

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!

A box moves 5\,\text m5m5, start text, m, end text horizontally when a force, F =10\,\text NF=10NF, equals, 10, start text, N, end text is applied at an angle, \theta =150\degreeθ=150°theta, equals, 150, degree as shown below. A dark grey box slides along a light grey horizontal surface. the box has a force vector F pulling up and to the left on it. A displacement vector d extends rightwards form the center of the box. The angle between vectors F and d is labeled theta. A dark grey box slides along a light grey horizontal surface. the box has a force vector F pulling up and to the left on it. A displacement vector d extends rightwards form the center of the box. The angle between vectors F and d is labeled theta. What is the work done on the box by FFF during the displacement?

Answers

Answer:

??

Explanation:

not completely sure hold on

Answer:

-43J

Explanation:

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