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2 years ago

9

Consider a large truck and a small car driving up a straight, steep hill. The truck is moving at 60 miles per hour and the car a

t 30 miles per hour. Assuming that the speeds are constant, which of the two vehicles experiences a larger net force?

1 answer:

tatiyna2 years ago

6 0

Answer:

Both the small car and the truck experience the same net force.

Explanation:

Net force can be defined as the vector addition of all the forces that follow up on a particle or a body.
It is a single force that changes the effect of the initial force on a particles motion.
If in case the net force acting on a particular body is zero then it means that the object or the body is not moving or accelerating at all and is in a state of stability.

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Multiplying or dividing vectors by scalars results in a. vectors. b. scalars. c. vectors if multiplied or scalars if divided. d.

klemol [59]

Let A = i+j+k be a vector and B = 3 be any scalar,
Multiplying A and B,
AB = (i+j+k)3 = 3i+3j+3k

Which is a new vector whose direction is same as the old but it's 3 times greater in length than the old vector(i+j+k).

Now, dividing A and B,
A/B = (i+j+k)/3 = $\frac{1}{3}i + \frac{1}{3}j + \frac{1}{3}k$

Which is again a new vector whose direction is same as the old but now it's 1/3 times small in length than the old vector.

Direction is same because we multiplied by positive scalar. If we multiply A by suppose -1, -4, -1000000 or any negative number, it's direction will reverse.

Thus, if we multiply a vector with scalar, it's length increases. If we divide, it shrinks.

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2 years ago

Read 2 more answers

The point on the graph that lies on the y-axis (vertical axis) is called the y-intercept. What does the y-intercept tell you abo

jekas [21]

Answer:

The starting position of the runner.

Explanation:

When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.

The runner starts at twenty, and ends at thirty.

Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.

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2 years ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago

(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp

12345 [234]

(a) $2.64\cdot 10^{-8} N$ north

We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is

$F=qvB sin \theta$

where

$q=0.500 \mu C = 0.500\cdot 10^{-6} C$ is the charge on the plane

v = 660 m/s is the velocity

$B=8.00\cdot 10^{-5} T$ is the magnitude of the magnetic field

$\theta=90^{\circ}$ is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

$F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N$

The direction can be found by using Fleming's left hand rule. We have:

- index finger: magnetic field direction (straight up)

- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

$W=mg=(80000)(9.8)=784,000 N$

As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.

8 0

2 years ago

This means that the speed at which the bullet travels across Earth's surface (its magnitude of horizontal velocity) does not aff

Dmitry_Shevchenko [17]

Answer: the speed at which it falls toward the Earth.

Explanation:

A bullet travelling across Earth's surface with some horizontal velocity is classical example of projectile motion.

Projectile motion is an idealization of the motion under the action of gravity neglecting the influence of the air (no drag force nor friction).

This kind of motion is the result of two independent motions: vertical motion and horizontal motion.

The observed net velocity is the vectorial sum of the vertical and horizontal velocities.

The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).

The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.

Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.

6 0

2 years ago