What is the weight of a 25 kg object on Earth with an acceleration due to gravity of 9.8m/s/s?

2.45 n
24.5 n
245 n
2450 n​

Answers

Answer 1
245 n hope this helps
What Is The Weight Of A 25 Kg Object On Earth With An Acceleration Due To Gravity Of 9.8m/s/s?2.45 N24.5

Related Questions

If an object is placed at distance of 16cm from a plane mirror, How far would it be from it's image?​

Answers

Explanation:

A plane mirror always creates an image with the same distance to the mirror as the object, only in the other direction. So both of them have a distance of 10cm, one is 10cm to the left, one 10cm to the right, thus their mutual distance is 20cm

You are riding in the passenger seat of a car as it goes around a tight turn. You slide across the seat to the passenger side door. Which statement below properly describes what is happening? a. You are exerting a centripetal force on the door. b. The door and seat are exerting a centripetal force on you that balances the centrifugal force of the turn. c. The car seat exerts a centripetal force on you, but not enough to keep you in place, so the door exerts the rest. d. The centrifugal force is greater than the force of friction between you and the seat, so you slide outward.

Answers

Answer:

Explanation:

answer C looks good

there isn't really a "centrifugal " force. :/    when we are pushed "back" in a car seat.. it's not because there is a force pushing us backwards... but a force pushing us forwards.. just like when turning too,   a force pushes us into the corner,  not a force pushing out of the corner.  :)  

When you are driving on the freeway and following the car in front of you, how close is too close? Let's do an estimation.
1. Pick a car model (preferably the one you drive, but can also be any car of your dream), and find its stopping distance at highway speeds (you can usually find this type of data online).
2. Assuming that the car in front of you suddenly does a hard brake. For simplicity, assume that its braking performance is about the same as yours. Then also assume a reasonable amount of reaction time on your part (the time delay between seeing the brake lights lit up and applying your own brake). In order for you not to run into the car your are following, what's the closest distance you need to keep between the two cars?
3. Redo the same calculation if the vehicle in front of you is a typical big-rig truck. Find its braking data online.
4. There is a rule of thumb which says that you must stay one car length behind the car in front of you for every 10 mi/h of driving speed. From your calculation, does this rule make sense?

Answers

Answer:

1) v= 90km/h  d = 70 m,  2)  x₁ = v t_r,  x₁ = 6.25 m, 3) x₁=6.25 no change

4) x = 22 m

Explanation:

1) for the first part, you are asked to find the minimum safety distance with the vehicle in front

The internet is searched for the stopping distance for two typical speeds on the highway

v (km/ h)      v (m/s)      d (m)

90                  25           70

100                 27.78      84

the safe distance is this distance plus the distance traveled during the person's reaction time, which can be calculated with infirm movement

              v = x / t_r

               x₁ = v t_r

the average reaction time is t_r = 0.25s for a visual stimulus and t_r 0.17 for an auditory stimulus

therefore the safe distance is

              x_total = x₁ + d

2) The distance is the sum of the distance traveled in the reaction

              x₁ = v t_r

for v = 90 km / h

              x₁ = 25 0.25

              x₁ = 6.25 m

for v = 100 km / h

              x₁ = 27.78 0.25

              x₁ = 6.95 m

the total distance is

               x_total = x₁ + d

for v = 90 km / h

             x_total = 25 0.25 + 70

             x_total = 76.25 m

this is the distance until the cars stop and do not collide

3) the stopping distance of a truck is

   v = 90 km / h       d = 100 m

in this case we see that the braking distance is much higher,

the safe distance is given by the distance traveled during the reaction, as the truck brakes slower than the car this distance does not change

4) let's analyze the empirical rule: maintain the length of a car for each increase in speed of v = 10 m / h = 4.47 m / s

for the car case at v = 90km / h = 25 m / s

according to this rule we must this to

           x = 25 / 4.47 = 5.6 cars

each modern car is about 4 m long so the distance is

           x = 22 m

we see that this distance is much greater than the reaction  distance so it does not make much sense

4. Name three examples of "concentrated" forms of energy.

Answers

Answer:

Nuclear power plant.

Gas stove.

Dam.

Gas pump.

Geothermal heat pump.

Power lines.

Solar panels.

Windmills.

Explanation:

Hope this helps :))

Answer:

gasoline,solar panels,geothermal heat pump,windmills

Explanation:

What x rays travel at the speed of

Answers

What is the speed? Like we need the pic or something

Two automobiles, each of mass 1000 kg, are moving at the same speed, 20 m/s, when they collide and stick together. In what direction and at what speed does the wreckage move (a) if one car was driving north and one south (b) if one car was driving north and one east?
A. The wreckage after collision is moving at the speed 18 m/s to the south.
B. The wreckage after collision is moving at the speed 9.0 m/s to the north.
C. The wreckage after collision is moving at the speed 9.0 m/s to the south.
D. The wreckage after collision is moving at the speed 18 m/s to the north.
E. The wreckage after collision is motionless.

Answers

Answer:

The reckage after collision is motionless (E)

Explanation:

The first law of thermodynamics states that energy is neither created nor destroyed but is converted from one form to another.

The kind of collision described in the question above is known as a perfectly inelastic collision, and in this type of collision, the maximum kinetic energy is lost because the objects moving in opposite directions have a resultant momentum that is equal, but in opposite directions hence they cancel each other out.

The calculation is as follows:

m₁v₁ + m₂v₂

where:

m₁ = m₂ = 1000kg

v₁ = 20 m/s

v₂ = -20 m/s ( in the opposite vector direction)

∴ resultant momentum = (1000 × 20) + (1000 × -20)

= 20000 - 20000 = 0

∴ The reckage after collision is motionless

Answer:

The wreckage after collision is moving at the speed 18 m/s to the south.

Explanation:

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