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3 years ago

1. How much force is needed to accelerate a 66 kg skier at 2 m/sec2?

Physics

1 answer:

Tju [1.3M]3 years ago

3 0

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 66 × 2

We have the final answer as

Hope this helps you

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Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

2 years ago

How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'

Serhud [2]

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity $(K.E.=\frac{1}{2}mv^2)$ . If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height $(P.E.=mgh)$ . If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.

4 0

3 years ago

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Aleonysh [2.5K]

Answer:

Explanation:

Objects with the same charge (Ex. North and north) repel each other.

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2 years ago

How are convergent boundaries different from divergent boundaries

fenix001 [56]

Convergent boundaries: the tectonic plates move toward each other

Divergent boundaries: the tectonic plates move away from each other

5 0

3 years ago

Two blocks connected by a light string are being pulled across a frictionless horizontal tabletop by a hanging 16.2-N weight (bl

Artemon [7]

Newton's second law allows us to find the results for the string tensions are:

T₁ = 6.7 N

T₂ = 16.54 N

Newton's second law gives a relationship between force, mass and acceleration of bodies

∑ F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration.

Free-body diagrams are representations of the forces applied to bodies without the details of them.

The reference system is a coordinate system with respect to which the forces decompose, in this case the x-axis is parallel to the plane and the positive direction in the direction of movement, the y-axis is perpendicular to the plane.

In the attachment we see a free-body diagram of the three-block system.

Let's apply Newton's second law to each body.

Block C

Y-axis

$W_c -T_2 = m_c a$

Block A

X axis

$T_2 - T_1 - W_a_x = m_a a$

Y axis

$N_a - W_a_y = 0$

Block B

X axis

$T_1 - W_b_x = m_b a$

Y axis

$N_b - W_b_y =0$

Let's use trigonometry to find the components of the weight.

Block A

cos θ = $\frac{W_a_y}{W_a}$

sin θ = $\frac{W_a_x}{W_a}$

$W_a_y = W_a cos \theta$

$W_a_x= W_a sin \theta$

Block B

cos θ = $\frac{W_b_y}{W_b}$

sin θ = $\frac{W_b_x}{W_b}$

$W_b_y = W_b cos \theta \\W_b_x = W_b sin \theta$

Let's write our system of equations.

$W_c - T_2 = m_c a \\ T_2 - T_1 - W_a_x = m_a a \\T_1 - W_b_x = m_b a$

Let's find the acceleration of the bodies, adding the equations.

$W_c - W_a_x - W_b_x = ( m_a+m_b+m_c) a\\$

The weight is

W = mg

Let's substitute

$(m_c - m_a -m_b ) g \ sin \theta = ( m_c+m_a+m_b) \ a \\a= \frac{ m_c-m_a-m_b }{ m_a+m_b+m_c} \ g sin \theta$

Indicate ma mass of the block a ma = 1.00 kg, the mass of the block b mb = 2.2 kg and the weight of the block c Wc = 16.2 N, let's find the mass of block c.

m_c = Wc / g

m_c = 16.2 / 9.8

m_c = 1.65 kg

we substitute the values

$a= \frac{1.65 -2.20 -1.00}{1.65+2.20+1.00} \ 9.8 \ sin \theta \\a= -0.3096 sin \theta$

The negative sign indicates that the system is descending, to be able to give a specified value an angle is needed, they assume that the angle of the ramp is 45º

a = - 0.3196 sin 45

a = -0.226 m / s

Taking the acceleration we are going to look for the tensions.

From the equation of block C

$W_c - T_2 = m_c a \\T_2 = m_c ( g-a)\\T_2 = 1.65 ( 9.8 + 0.226)$