If an object is not accelerating what can you determine about the sum of all the forces on the object

Phil is riding a scooter and pushes off the ground with his foot. this causes him to accelerate at 12 m /s. Phil weighs 600 N. h

Dvinal [7]

Answer:

734.16 kg m/ $s^{2}$

Explanation:

The problem is asking for the Force of pushing off the ground.

The formula of Force is: F = mass x acceleration

Given = Mass: 600 newtons (N)

Acceleration: 12 m/ $s^{2}$

We have to convert the mass into kg first. Remember that 1 kg is equal to 9.80665 newtons.

Let x be the mass in newtons.

Let's convert: $\frac{1 kg}{9.80665 N}$ x $\frac{x}{600 N}$ = $\frac{600}{9.80665}$ = 61.18 kg

Phil's weight is 61.18 kg

Let's go back to finding the force.

F = m x a

F = 61.18 kg x 12 m/ $s^{2}$

F = 734.16 kg m/ $s^{2}$

7 0

3 years ago

What happens when a negatively charged object A is brought near a neutral object B?

Rashid [163]

The answer is E. The objects loses all its charge

6 0

3 years ago

Choose the items that help to fully describe voltage.

Zina [86]

Answer:

the easy way to describe this is to use a light as an example.

Explanation:

Voltage is pretty much the loop used to help use a lightbulb to emit light. Without voltage, we would be unable to use lightbulbs. This applies to much more than a lightbulb, but it's the easiest way to describe how voltage works.

8 0

2 years ago

In one cycle, a freezer uses 800 J of electrical energy in order to remove 1735 J of heat from its freezer compartment at 10.0°F

Lubov Fominskaja [6]

Answer:

Explanation:

A)

W = work done by the freezer = 800 J

Q = heat removed from the freezer = 1735 J

Q' = Heat expelled into the room

Coefficient of performance is given as

$\beta = \frac{Q}{W}$

inserting the values

$\beta = \frac{1735}{800}$

$\beta = 2.2$

B)

Heat expelled is given as

Q' = W + Q

Q' = 800 + 1735

Q' = 2535 J

8 0

3 years ago

Assuming this is a distance time graph( ignore the speed time title) assume metres on vertical scale. describe in as much detail

riadik2000 [5.3K]

Answer:

The journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance travelled during the journey from the start point A to the final point B is 40 m

Explanation:

From the start point A to point B, we have;

The speed from A to B = 10 m/(10 s) = 1 m/s

The distance traveled from A to B = 10 m

The time it takes to move from A to B = 10 seconds

From the point B to point C, we have;

The distance traveled from B to C = 0 m, (stationary)

The time it remains at point B distance from the start point = 10 seconds

The speed between point B to C = 0 m/(10 s) = 0 m/s

From the point C to point D, we have;

The distance traveled from C to D = 10 m

The time it takes to move from C to D = 5 seconds

The speed between point C and D = 10 m/(5 s) = 2 m/s

From the point D to point E, we have;

The distance traveled from D to E = 0 m, (stationary)

The time it remains at point D distance from the start point = 10 seconds

The speed between point D to E = 0 m/(10 s) = 0 m/s

From the point E to point F, we have;

The distance traveled from E to F = 20 m (return journey starts at point E)

The time it takes to move from E to F = 5 seconds

The speed between point E to F = 20 m/(5 s) = 4 m/s (Return journey)

Therefore, the journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance moved, 'd', to and from the start point with reference to the graph is given as follows;

d = (From A to B) 10 m + (From B to C) 0 m + (From C to D) 10 m + (From D to E) 0 m + (From E to F) 20 m = 40 m

The total distance travelled in the journey is 40 m

The total displacement, $\underset{d}{\rightarrow}$ = 10 m + 10 m - 20 m = 0 m

7 0

3 years ago