If the force on an object is negative, what is known about the change in velocity?

Pleaseee someone help me TEST TOMORROW

Liono4ka [1.6K]

3.A

4.c

5.A
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4 0

3 years ago

Read 2 more answers

An acorn falls from an oak tree. You note that it

Vaselesa [24]

Answer:9.8 m/s²

Explanation:

It was going at 9.8m/s² as this is the acceleration of an object due to gravity

when an object falls it accelerates at a consant and uniform speed which is 9.8m/s²

6 0

3 years ago

Farmer Tamar airlifted her sheep from 1000 meters to 1,200 meters

attashe74 [19]

Question

What was the change in potential energy of the sheep?

Answer:

1962(m) J

Explanation:

Potential energy is given by

PE=mgh

Change in potential energy is given by

∆PE=mg(h2-h1)

Where ∆PE is the change in potential eneegy, g is acceleration due to gravity, h2 is final height and h1 is the initial height.

Substituting 9.81 m/s² for g, 1200 m for h2 and 1000 m for h1 then

∆PE=m*9.81(1200-1000)=1962m Joules

Since we are not given the mass of rhe sheep, the answer will be 1962(m) Joules

4 0

3 years ago

A ball that has a mechanical energy of 65 J has 12 J of kinetic energy. The ball has

bulgar [2K]

Answer:

The ball has a potential energy of 53 J.

Explanation:

Mechanical energy, E = Kinetic energy + Potential energy

E = K.E + P.E

65 = 12 + P.E

P.E = 65 – 12

P.E = 53 J

Therefore the potential energy of the ball is 53 J

8 0

3 years ago

In a hydraulic lift, if the pressure exerted on the liquid by one piston is increased by 100 N/m2 , how much additional weight c

shepuryov [24]

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

$P = \frac{F}{A}$

Hydraulic Lift - After change

$P + \Delta P = \frac{F + \Delta F}{A}$

Where:

$P$ - Hydrostatic pressure, measured in pascals.

$\Delta P$ - Change in hydrostatic pressure, measured in pascals.

$A$ - Cross sectional area of the hydraulic lift, measured in square meters.

$F$ - Hydrostatic force, measured in newtons.

$\Delta F$ - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

$\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}$

$\Delta P = \frac{\Delta F}{A}$

$\Delta F = A\cdot \Delta P$

Given that $\Delta P = 100\,Pa$ and $A = 25\,m^{2}$ , the additional weight is:

$\Delta F = (25\,m^{2})\cdot (100\,Pa)$

$\Delta F = 2500\,N$

The additional mass needed for the additional weight is:

$\Delta m = \frac{\Delta F}{g}$

Where:

$\Delta F$ - Additional weight, measured in newtons.

$\Delta m$ - Additional mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

If $\Delta F = 2500\,N$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }$

$\Delta m = 254.92\,kg$

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

3 0

3 years ago