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3 years ago

A scalar quantity has

1 answer:

3 0

A scalar quantity is described by a magnitude only. It includes speed, time, mass, and volume. There is no direction involved. Therefore, your answer is C- Magnitude but no direction.

I hope I've helped! :)

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A 3,220 lb car enters an S-curve at A with a speed of 60 mi/hr with brakes applied to reduce the speed to 45 mi/hr at a uniform

Grace [21]

The magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. Its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

The given data in the problem is;

The weight is,W= 3,220 lb

The speed is,u= 60 mi/hr

The reducing speed is,v= 45 mi/hr

The distance traveled is,d= 300 ft

The radius of curvature of the path of the car at B is,R= 600 ft.

1 mile = 5280 ft

From the Newtons' equation of motion;

$\rm v^2 = u^2 +2ad \\\\ \rm (45 \times \frac{5280}{3600} )^2 = (60 \times \frac{5280}{3600} )^2 +2a\times 300 \\\\$

The tangential accelerations are;

$\rm a_t = \frac{66^2 -88^2}{600} \\\\ \rm a_t = -5.65 ft/sec^2 \\\\$

The force is found as;

$\rm \sum F = ma \\\\ \frac{3220}{32.2} \times -5.65 \\\\ F_T= 565 \ lb$

The normal force is;

$\rm F_n = \frac{3220}{32.2} \times \frac{66^2}{600} \\\\ F_N =726 \ lb$

The net or the total friction force exerted by the road on the tires at B. is found as;

$\rm F = \sqrt{(565)^2+(726)^2} \\\ F = 919.946 \ lb$

Hence, the magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

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6 0

1 year ago

Help plz I’ll mark brainliest

marin [14]

I think it’s concave

3 0

2 years ago

Read 2 more answers

Exactly one pound of bread dough is placed in a baking tin. The dough is cooked in an oven at 350°F, releasing a wonderful aroma

Morgarella [4.7K]

Answer:

The mass of the baked loaf will be less than the dough.

Explanation: When heat is applied to food substance or products like the one pound the substance or material gains a higher temperature, the increase in temperature causes moisture inherent or added to the product in this case the one pound dough to be lost, the one pound dough prepared at room temperature, once it is placed inside the oven at 350 degrees Fahrenheit it will lose moisture in the form of vapor to the environment as noticed in the aroma, the moisture lost will eventually reduce it mass/weight (kilograms or grams) by some percentage or quantities(kilograms or grams)

7 0

3 years ago

A 300-kg piano being held by a crane is accidentally dropped from a height of 15 meters. a. What is the speed of the piano just

FinnZ [79.3K]

Answer:

a) 17.16m/s

b) 44,145J

c) Sound the piano makes when hitting the ground, vibration of the ground, heat.

d) i) It's smaller due to the energy dissipated by the friction between air and the parachute.

ii) It stays the same, the only difference is that the dissipated energy is distributed between air resistance and the kinetic energy dissipated by the ground whent he piano hits it.

Explanation:

In order to solve this problem we must start by doing a drawing of the situation, which will help us visualize the problem better. (See attached picture).

So, in this problem we can ignore air resistance so we can say that the energy is conserved, this is the total initial energy is the same as the total final energy, so we get that:

$U_{0}+K_{0}=U_{f}+K_{f}$

When the piano is released it has an initial speed of zero, so the initial kinetic energy is zero. When the piano hits the ground it will have a height of 0m, so the final potential energy is zero as well. This will simplify our equation:

$U_{0}=K_{f}$

We know that potential energy is given by the formula:

U=mgh

and kinetic energy is given by the formula:

$K=\frac{1}{2}mv^{2}$

which can be substituted in our equation:

$mgh=\frac{1}{2}mv^{2}$

we can divide both sides of the equation into the mass of the piano, so we get:

$gh=\frac{1}{2}v^{2}$

which can be solved for the final velocity which yields:

$v=\sqrt{2gh}$

we can now substitute the data provided by the problem so we get:

$v=\sqrt{2(9.81m/s^{2})(15m)}$

which yields:

v=17.16m/s

Since energy is conserved, this means that the total dissipated energy will be the same as the potential energy, so we get that:

E=mgh

$E=(300kg)(9.81m/s^{2})(15m)$

which yields:

E=44,145J

When the piano hits the ground, the kinetic energy it had will be transformed to other types of energy, mostly vibration and heat. The vibration will turn to sound due to the movement of air created by the piano itself and the ground. And heat is created by the friction between the molecules created by the vibrations and the collition itself. So some of the indicators of this release of energy could be:

-Sound

-Vibration

-Heat.

i) The amount of inetic energy dissipated would decrease due to the friction between air and the parachute. Since air is resisting the movement of the piano, this will translate into a loss of energy, if we did an energy balance we would get that:

$U_{0}=K_{f}+E_{p}$

The total amount of energy is conserved but it will be distributed between the energy lost due to air resistance and the kinetic energy the piano has at the time it hits the ground.

ii) So the total amount of energy dissipated remains the same, the only difference is that it will be distributed between air resistance and the kinetic energy of the piano.