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Oliga [24]
3 years ago
7

Two arrows are shot vertically upward. the second arrow is shot after the first one, but while the first is still on its way up.

the initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. suppose that the initial speed of the first arrow is 31.8 m/s and that the second arrow is fired 1.82 s after the first. determine the initial speed of the second arrow.
Physics
1 answer:
Neko [114]3 years ago
7 0
<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get....... 17.15 m/s</span>
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Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

Torque = \vec{r} \times \vec{F}

Where \vec{r} is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), \vec{F} is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench.  By doing this we have that r (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

35cm * \dfrac{1m}{100cm} = 0.35m

Now using the torque formula:

Torque = \vec{r} \times \vec{F}

Torque = rFsin(\theta)}

Where \theta is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector  \theta = 90°, thus :

Torque = rFsin(\theta)}

65 N m = (0.35m)Fsin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

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3 years ago
Using the strap at an angle of 31.0° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15.0
lora16 [44]

Answer:

<h2>154.73N</h2>

Explanation:

The question is incomplete. Here is the complete question.

Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.

Check the diagram related to the question in the attachment below for better understanding.

The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.

The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).

Ty = 15sin31°

Ty = 7.73N

W = mass * acceleration due to gravity

W = 15.0*9.8

W = 147N

The normal force is therefore expressed as;

N = Ty + W

N = 7.73 + 147

N = 154.73N

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