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3 years ago

In right ΔPQR, sin P = 3 over 5 , and cos P = 4 over 5. What is tan P?

1 answer:

5 0

Not sure if you learned S=O/H,C= A/H, T=O/A, but Sin is equal to opposite over hypotenuse, so draw a triangle, and label those angles (opposite= 3 and hypotenuse=5). Also, Cos is equal to adjacent over hypotenuse, so label those as well (you already have hypotenuse from Sin, and the Adjacent angle = 4). Therefore, Tan which equals opposite over adjacent is 3/4

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In gym class you run 22 m horizontally, then climb a rope vertically for 6.2 m. What is the direction angle of your total displa

miskamm [114]

Im 99% sure
tan^-1 (6.2/22)= 15.7º
with 2 significant figures 16º

7 0

3 years ago

Read 2 more answers

Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support

alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

$\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)$

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

$r_{M\perp}(F_M)-r_{W\perp}(W)=0$

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

$r_{M\perp}(F_M)=r_{W\perp}(W)$

Now, we divide by the distance of the muscle:

$(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}$

Now, we substitute the values:

$F_M=\frac{(2.4cm)(50N)}{5.1cm}$

Now, we solve the operations:

$F_M=23.53N$

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

$\Sigma F_v=F_j-F_M-W$

Since there is no vertical movement the sum of vertical forces is zero:

$F_j-F_M-W=0$

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

$F_j=F_M+W$

Now, we plug in the values:

$F_j=23.53N+50N$

Solving the operations:

$F_j=73.53N$

Therefore, the force is 73.53 Newtons.

8 0

1 year ago

Suppose two carts, one twice as massive as the other, fly apart when the

disa [49]

Answer:

the question this morning

Explanation:

90+70 look than

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3 years ago

If an astronaut had a mass of 30 kg on the moon what would his mass be on earth?

Novay_Z [31]

Explanation:

A one-kilogram mass is still a one-kilogram(as mass is an intrinsic property of the object) but the downward force due to gravity, and therefore it's weight, is only one-sixth of what the object would have on the Earth. So man of mass 180 pounds weights only about 30 pounds-force when visiting the moon

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Have a nice day

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3 years ago

A woman can row her canoe at 2.5 m/s. If she faces an opposing current of 3.0 m/s, how fast will she go forward?

taurus [48]

Answer:

V = - 0.5 [m/s]

Explanation:

In order to solve this problem, we must use the principle of relative speeds. This is for an observer who is on the edge of the river he can see how the river moves to the left and the woman tries to move to the right but can not since:

$V_{total}=-3+2.5\\V_{total}=-0.5 [m/s]$

That is, the person sees how the woman moves to the left but with avelocity of 0.5 [m/s] to the left

8 0

3 years ago