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3 years ago

Throw a rock horizontally at 22 m/s from a 315 m tall building. How far from the base of the building will it land?

Physics

1 answer:

creativ13 [48]3 years ago

6 0

Answer:

176.44 m

Explanation:

y = (vertical vi)t + (1/2)gt^2

315 = 0 + (1/2)(9.8)t^2

t = 8.02s

x = (horizontal vi)t + (1/2)at^2

x = (22)(8.02) + 0

x = 176.44 m

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A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas

Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

$\varepsilon' =\dfrac{S_y}{E}$

$\varepsilon' =\dfrac{240}{110\times 1000}$

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

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3 years ago

What unit is used to count atoms and molecules?

murzikaleks [220]

the answer in my opinion would be A

7 0

3 years ago

A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t

rusak2 [61]

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

$v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt$

v_ave = v0+a(T/2)

We can esaily note that when t=T/2

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

$x(t) = x_A + v_0 t + \frac{1}{2}at^2$

Where x_A is the position of point A. Therefore, the car will be at:

x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2

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2 years ago

You can comfortably hold your fingers close beside a candle flame, but not very close above the flame. why? challenge (optional)

Inessa05 [86]

The candle flame releases hot gases, which directly go in upwards directions. Due to which the air near the flame of the candle is very hot and dense. The particles along with vapour move up. And since the sideways, the air is not very dense and hot, we are able to hold the candle. In anti-gravity region, there will be no density differences and also, the convection process wont occur. So, the candle quickly snuffs off.

6 0

3 years ago

Read 2 more answers

A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient

gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0

2 years ago

Throw a rock horizontally at 22 m/s from a 315 m tall building. How far from the base of the building will it land?​

Throw a rock horizontally at 22 m/s from a 315 m tall building. How far from the base of the building will it land?