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2 years ago

A car accelerates uniformly from rest at a speed of 1.67 ft s^2 over a distance of 5 yards.What is the acceleration of a car?

Physics

2 answers:

stepladder [879]2 years ago

7 0

Answer:

a= 0.256 ft/s^2

Explanation:

Solution:

Since the car has started from rest the initial velocity will be zero

i.e. Vi=0

It gets a speed of 1.6 fts^2,

i.e. Vf= 1.6 fts^2

The distance is 5 yards

i.e d= 5 yards

According to the equation of motion

2as= VF^2- Vi^2

Thus a= (VF^2- Vi^2)/2s

a= (1.6^2 - 0)/ 2*5

a= 0.256 ft/s^2

Nina [5.8K]2 years ago

5 0

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

acceleration= 23/1.30

accelaration= 17.69 m/s^2

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Among three bases, x−, y−, and z−, the strongest one is y−, and the weakest one is z−. rank their conjugate acids, hx, hy, and h

mr_godi [17]

Bases:
x- middle strength
y- strongest base
z- weakest base

The bases and conjugate acids have an inverse strength. The stronger the conjugate base, the weaker its conjugate acid.

I am considering two scenarios:
Scenario 1:
y- strongest base ; hy - weakest conjugate acid
x- middle base ; hx - middle conjugate acid
z- weakest base ; hz - strongest conjugate acid

hz → hx → hy

Scenario 2:
y-strongest base ; hy - weakest conjugate acid
x-middle base ; hx - strongest conjugate acid
z-weakest base ; hz - middle conjugate acid

hx → hz → hy

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2 years ago

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Pavlova-9 [17]

Answer:

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1 year ago

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

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2 years ago

Which statement is true about a planet’s orbital motion?

lana66690 [7]

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

$V = \sqrt{\frac{GM}{R + h} }$

Where

V - velocity of the orbit at a height h from the surface

R - Radius of the second object

G - Gravitational constant

h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

$1/2 mV^{2} = GMm/R$

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

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3 years ago