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3 years ago

How much work is accomplished when a force of 300N pushes a box across the floor for a distance of 100 meters?

Physics

2 answers:

Nesterboy [21]3 years ago

5 0

So the correct ans is B.

hope it helps u.

Wewaii [24]3 years ago

4 0

Answer:

D.30000 J

Explanation:

Work done = force x distance

From the question

Force F = 300N

Distance d = 100m

Therefore using the above formula, we have

W.d = 300N x 100m

= 30000 J

Work done is measured in Joules J

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A projectile lands at the same height from which it was launched. which initial velocity will result

Serhud [2]

The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin θ.

The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.

If the air resistance is ignored, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.

Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

Learn more here; brainly.com/question/12870645

5 0

3 years ago

If the mass of the block is 5 kg and the speed 7 m/s, what is the work done on the block?

Kamila [148]

Answer:

A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms

2 ).

Mass of the block=5kg

Coeffecient of friction=0.2

external applied force, F=40N

The angle at which the force is applied=30degree

So the horizontal component of force=Fcos30=40×

23 =20 3 N

While the uertical component of the force acting in upward direction=Fsin30=40× 21

=20N

The normal reaction from the surface (N)=mg−Fsin30=50−20=30N

So the ualue of limiting friction=μN=0.2×30=6N

Hence the net horizontal force on the block=Fcos30=μN=20

N−6N=28.64N

The horizontal acceleration of the block=

Fcos30−μN = 528.64

=5.73m/s 2

8 0

3 years ago

Multiply. (2x + 4)(x - 4)

sergij07 [2.7K]

Answer:

(2x + 4)(x - 4)=2x^2-4x-16

7 0

2 years ago

A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has

iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
So, we can write the following general equation, taking the initial and final values of the energies:

$\Delta K + \Delta U = W_{ffr} (1)$

Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
⇒ Kf = 1/2*m*vf² (2)
The change in the potential energy, can be written as follows:

$\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

where k = force constant = 815 N/m

xf = final displacement of the block = 0.01 m (taking as x=0 the position

for the spring at equilibrium)

x₀ = initial displacement of the block = 0.03 m

Regarding the work done by the force of friction, it can be written as follows:

$W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
Fn = Fg= m*g (5)
Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

$\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$

$\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)$

Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

$\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)$

$v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)$

7 0

3 years ago

Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo

sergiy2304 [10]

Answer:

$a=9\ cm/s^2$

Explanation:

Average Acceleration

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that.

When $t_2=4\ sec$ , the graph shows a value of $v_2=36\ cm/s$