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ki77a [65]
4 years ago
15

This is junior year english

Physics
2 answers:
pychu [463]4 years ago
6 0
D. is the correct answer.
Alla [95]4 years ago
6 0
I think the answer is D
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a horizontal force of 100N is required to push a crate across a factory floor at a constant speed. What is the net force acting
Bas_tet [7]

If the crate is moving along the floor in the same direction with a constant speed, it is in dynamic equilibrium. Equilibrium means there is no net force acting on the crate.

Since there is no net force on the crate, there must be a friction force on the crate equal in magnitude and opposite in direction to the applied horizontal force. Therefore the force of friction acting on the crate is 100N.

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How much of the earth's animal life exists in the oceans?<br><br> 40%<br> 60%<br> 80%<br> 99%
Alexxx [7]

How much of the earth's animal life exists in the oceans?

80% is the answer.

7 0
3 years ago
Why does a curve on a distance-time graph indicate acceleration? an increasing amount of time passes for each unit of distance a
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because speed can change overtime for a car, and isn't always constant.

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3 years ago
Read 2 more answers
a car with a mass of 1200 kilograms is moving around a circular curve at a uniform velocity of 20 m/s the centripetal force on t
jek_recluse [69]
Coupla things wrong with this question, Sam.
Let's clean those up first, and then we'll work on the answer.

-- The car is NOT moving with uniform velocity.
'Velocity' includes both speed and direction.  If either of these
changes, it's a change of velocity.
On a circular track, the car's direction is CONSTANTLY changing,
so its velocity is too. 
The thing that's uniform is its speed, not its velocity.

-- A 'neutron' is a subatomic particle found in the nucleus of most
atoms.  It's not a unit of force.  The unit of force is the 'Newton'.
_______________________

OK. A centripetal force of 6,000 newtons keeps 1,200 kg of mass
moving in a circle at 20 m/s.

The formula:

                                     Centripetal force = (mass) (speed)² / (radius)

Multiply each side
by 'radius':                (centripetal force) x (radius) = (mass) x (speed)²

Divide each side by
'centripetal force':       Radius = (mass) x (speed)² / (centripetal force)

Write in the numbers
that we know:              Radius = (1200 kg) (20 m/s)² / (6000 Newtons)

                                                 = (1200 kg) (400 m²/s²) / (6000 Newtons)

                                                 = (480,000 kg-m²/s²) / (6000 kg-m/s²)

                                                 = (480,000 / 6000) meters
                                                
                                                 =         80 meters .   
8 0
3 years ago
Read 2 more answers
Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

a=1.5 m/s^2

Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

x=ut+\frac{1}{2}at^2

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0
3 years ago
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