All categories

3 years ago

If you are pushing a table with a force of 5 newtons what is the force pushing back on you

1 answer:

4 0

The force pushing back to you would be 5 newtons as Newtons 3rd law says that every action force has an opposite but equal reaction force acting upon it (if i remember correctly )

You might be interested in

a motorcycle accelerates from 15 m/s to 20 m/s over a distance of 50 meters. what is its average acceleration?

devlian [24]

For this, you need the v-squared equation, which is v(final)² = v(initial)² + 2aΔx The averate acceleration is thus a = (v(final)² - v(initial)²) / 2Δx = (20² - 15²) / 2(50) = 175 / 100 = 1.75 m/s² So the average acceleration is 1.75 m/s²

5 0

3 years ago

A miner finds a small mineral fragment with a volume of 5.74 cm^3 and a mass of 28.7 g. What is the density of that mineral frag

statuscvo [17]

P=M÷V
p=28.7÷5.74
p=5

The density is 5.

8 0

3 years ago

Read 2 more answers

What does a animal and a plant cell have in common.

Allushta [10]

Answer:

Structures that are common to plant and animal cells are the cell membrane, nucleus, mitochondria, and vacuoles. Structures that are specific to plants are the cell wall and chloroplasts.

Explanation:

if correct plzz brainliest :D

8 0

2 years ago

Read 2 more answers

Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right c

masya89 [10]

Answer:

vₓ = xg/2y

Explanation:

In this question, let us find the time it takes for the ball on the right that has zero initial velocity to reach the ground.

By newton equation of motion we know that

y = v₀ t - ½ g t²

t = 2y / g

This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance

vₓ = x/t

vₓ = xg/2y

Where we assume that x and y are known.

7 0

3 years ago

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

Rudik [331]

Answer:

1) P₁ = -2 D, 2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

$\frac{1}{f_1} = \frac{1}{q}$

q = f₁

2) for an object located at p = 25 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}$

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}$

$\frac{1}{f_2}$ = 0.06

f₂ = 16.67 cm

the expression for the power of the lenses is

P = $\frac{1}{f}$

where the focal length is in meters

1) P₁ = 1/0.50

P₁ = -2 D

2) P₂ = 1 /0.16667

P₂ = 6 D

4 0

2 years ago