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harkovskaia [24]

3 years ago

14

Suppose you walk 17 m straight west and then 26.5 m straight north. What is the compass direction, in degrees measured West of N

orth, of a line connecting your starting point to your final position

1 answer:

garik1379 [7]3 years ago

6 0

Answer:32.7

Explanation:

You must first make a right triangle with the two legs being 17 m west and 26.5 m north. To find the degree measurement west of north you must find the angle between the north leg and the hypotenuse. then just use inverse tangent of (17/26.5) which is 32 .7 degrees

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Why is the mass of an atom's electrons not included in the atom's mass number?

Katen [24]

Answer: A: the electrons move so fast that their mass cannot be measured

Explanation: because they are so fast scientist and other people cannot measure them

8 0

3 years ago

A worker exerts a pulling force on a box. The worker exerts this force by attaching a rope to the box and pulling on the rope so

shusha [124]

Answer:

<em>The magnitude of the pulling force is 20.66% of the gravitational force acting on the box</em>

Explanation:

<u>Accelerated Motion</u>

The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.

Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by

$\displaystyle x=\frac{at^2}{2}$

Solving for a

$\displaystyle a=\frac{2x}{t^2}$

$\displaystyle a=\frac{2\cdot 10}{5.12^2}$

$a=0.763\ m/s^2$

Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.

The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:

$T_y+N=m.g$

There Ty is the vertical component of the tension of the rope, N is the normal force, and m is the mass of the box

The decomposition of T gives us

$T_y=Tsin\theta$

$T_x=Tcos\theta$

Solving the above equation for N

$Tsin\theta+N=m.g$

$N=m.g-Tsin\theta\text{..........[1]}$

Now for the x-axis, there are two forces acting on the box, the x-component of the tension and the friction force Fr. Those forces are not equilibrated, thus acceleration is produced:

$Tcos\theta-F_r=m.a$

Recalling that

$F_r=\mu N$

$Tcos\theta-\mu N=m.a$

Replacing N from [1]

$Tcos\theta-\mu (m.g-Tsin\theta)=m.a$

Operating

$Tcos\theta-\mu m.g+\mu Tsin\theta=m.a$

Solving for T

$T(cos\theta+\mu sin\theta)=m.a+\mu m.g$

$\displaystyle T=\frac{m.a+\mu m.g}{cos\theta+\mu sin\theta}$

$\displaystyle T=m\frac{a+\mu g}{cos\theta+\mu sin\theta}$

We don't know the value of m, thus we'll plug in the rest of the data

$\displaystyle T=m\frac{0.763+0.10\cdot 9.8}{cos36.8^o+0.10 sin36.8^o}$

$T=2.0252m$

Dividing by the weight of the box m.g

$T/(m.g)=2.0252/9.8=0.2066$

Thus, the magnitude of the pulling force is 20.66% of the gravitational force acting on the box

7 0

3 years ago

A frog jumps vertically upward from a 20m tall building with an initial velocity of 8.1m/s. How high above the ground will the f

topjm [15]

Answer:

Explanation:

Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.

Therefore,

u = 10 m/s, initial upward velocity.

H = - 20 m, position of the ground.

g = 9.8 m/s², acceleration due to gravity.

Part (a)

When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore

u² - 2gh = 0

h = u²/(2g) = 10²/(2*9.8) = 5.102 m

At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.

Answer: 25.1 m above ground

Part (b)

Let v = the velocity when the frog hits the ground. Then

v² = u² - 2gH

v² = 10² - 2*9.8*(-20) = 492

v = 22.18 m/s

Answer: The frog hits the ground with a velocity of 22.2 m/s

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A 15.0 m long steel rod expands when its temperature rises from 34.0 degrees C to 50.0 degrees C. What is the change in the beam

azamat

0.00288................

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Streamlined means less air resistance. The car will need less force so less fuel in order to overcome air resistance. This will make it more fuel efficient.

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