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3 years ago

C) m1: D) m2: E) r. Please help me I need help now

Physics

1 answer:

Nesterboy [21]3 years ago

6 0

A) F = gravitational force

b) G = universal gravitational constant (6.67 × 10-11 N-m2/kg2)

c) m1 = mass of the body 1

d) m2 = mass of body 2

e) r = radius or distance between the two bodies.

Hope this helps!

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A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come

Yuliya22 [10]

Answer:

(a) vo = 24.98m/s

(b) t = 5.09 s

Explanation:

(a) In order to calculate the the initial speed of the ball, you use the following formula:

$y=y_o+v_ot-\frac{1}{2}gt^2$ (1)

y: vertical position of the ball = 2.44m

yo: initial vertical position = 0m

vo: initial speed of the ball = ?

g: gravitational acceleration = 9.8m/s²

t: time on which the ball is at 2.44m above the ground = 5.00s

You solve the equation (1) for vo and replace the values of the other parameters:

$v_o=\frac{y-y_o+1/2gt^2}{t}$

$v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}$

The initial speed of the ball is 24.98m/s

(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:

$0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s$

The time that the ball takes to arrive to the ground is 5.09s

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3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$