PHYSICS COLLEGE

A bridge is made with segments of concrete 91 m long (at the original temperature). If the linear expansion coefficient for concrete is 1.2 × 10−5 ( ◦C)−1 , how much spacing is needed to allow for expansion for an increase in temperature of 56◦F? Answer in units of cm.

Answers

Answer 1

Answer:

Answer:

change in length is 3.397 cm

Explanation:

Given data

long = 91 m = 9100 cm

coefficient for concrete (a) = 1.2 × 10−5 ( ◦C)−1

temperature = 56 F = (56× 5/9) ◦C

to find out

how much spacing is needed to allow

solution

we know allow space is given by this formula

change in length = coefficient for concrete × given length × temperature .............1

put all value in equation 1

change in length = 1.2 × 10−5 × 9100 × (56× 5/9)

change in length = 3.397 cm

so change in length is 3.397 cm

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A revolutionary war cannon, with a mass of 2260 kg, fires a 21 kg ball horizontally. The cannonball has a speed of 105 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?

Answers

Answer:

0.97566 m/s

Explanation:

$m_1$ = Mass of cannon = 2260 kg

$v_1$ = Velocity of cannon

$m_2$ = Mass of ball = 21 kg

$v_2$ = Velocity of ball = 105 m/s

As the momentum of the system is conserved we have

$m_1v_1=m_2v_2\n\Rightarrow v_1=(21* 105)/(2260)\n\Rightarrow v_1=0.97566\ m/s$

The velocity of the cannon is 0.97566 m/s

A hollow sphere of radius 0.25 m is rotating at 13 rad/s about an axis that passes through its center. the mass of the sphere is 3.8 kg. assuming a constant net torque is applied to the sphere, how much work is required to bring the sphere to a stop?

Answers

The work required to bring the sphere to stop is equal to the kinetic energy possessed by the sphere.

Kinetic energy of a rotating body is given by,

K.E = $(1)/(2)Iw^(2)$

Here, I= Moment of inertia of hollow sphere,

Since, the hollow sphere is rotating about the axis passing through its center, I = $(2)/(3)MR^(2)$

M= Mass of the sphere= 3.8 kg,

R= Radius of gyration= Radius of the sphere= 0.25 m

w= Angular speed of the sphere = 13 rad/s

Substituting the values,

Kinetic energy = $(1)/(2) *(2)/(3) (3.8)(0.25)^(2)(13.0)^(2)$

= 13.4 J

∴ Work required to bring the sphere to stop is 13.4 J.

A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:

Answers

Answer:

The ratio is $(RE)/(TE) = (2)/(3)$

Explanation:

Generally the Moment of inertia of a spherical object (shell) is mathematically represented as

$I = (2)/(3) * m r^2$

Where m is the mass of the spherical object

and r is the radius

Now the the rotational kinetic energy can be mathematically represented as

$RE = (1)/(2)* I * w^2$

Where $w$ is the angular velocity which is mathematically represented as

$w = (v)/(r)$

=> $w^2 = [(v)/(r)] ^2$

$RE = (1)/(2)* [(2)/(3) *mr^2] * [(v)/(r) ]^2$

$RE = (1)/(3) * mv^2$

Generally the transnational kinetic energy of this motion is mathematically represented as

$TE = (1)/(2) mv^2$

$(RE)/(TE) = ((1)/(3) * mv^2)/((1)/(2) * m*v^2)$

$(RE)/(TE) = (2)/(3)$

A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam

Answers

Answer:

the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Explanation:

Given the data in the question;

initial velocity; u = 0 m/s

height; h = 2.5 m

we find the velocity of the ball just before it touches the foam.

using the equation of motion;

v² = u² + 2gh

we know that acceleration due gravity g = 9.81 m/s²

so we substitute

v² = ( 0 )² + ( 2 × 9.81 × 2.5 )

v² = 49.05

v = √49.05

v = 7.00357 m/s

Now as the ball touches the foam

final velocity v₀ = 0 m/s

compresses S = 3 cm = 0.03 m

v₀² = v² + 2as

we substitute

( 0 )² = 49.05 + 0.06a

0.06a = -49.05

a = -49.05 / 0.06

a = -817.5 m/s²

Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Dawn is trying to find out how much weight she can push across the room. She is really trying to find her

Answers

Dawn is trying to figure out how much weight she can push with her strength, or what her maximum pushing force is, across the room. She could do an experiment to find out.

She must first prepare a testing space with a flat, smooth surface to reduce friction. She can then progressively add weights to a cart or other object and use all of her strength to try to push it across the room. She can determine her maximum pushing force by noting the heaviest weight she can move. For a variety of jobs, including moving furniture or participating in physical sports that call for pushing heavy things, this knowledge can be essential.

To know more about physical sports, here

brainly.com/question/19595632

#SPJ2

Answer:

Muscular strength

Explanation:

She is testing her strength while pushing the weights

What allows two or more atoms to be held together? shared protons shared electrons shared energy shared neutrons

Answers

It is shared electrons.

The following information should be considered:

In the case when two or more atoms can be together at the time when they share electrons with each other.
By sharing, they create a covalent bond and that way the atoms can be stable.

learn more: brainly.com/question/2514933?referrer=searchResults

Answer:

try electrons i hope this helps!!

Explanation:

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