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iris [78.8K]
3 years ago
5

How do you calculate time given the acceleration and distance?​

Physics
1 answer:
mart [117]3 years ago
7 0

Explanation:

acc is velocity over time and the velocity rule is distance over time ..so acc would be distance over time power2

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An electrical insulator has
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A vessel containing 200 ml hydrogen gas at pressure 500torr at temperature 10 degree C.Find its volume when pressure increased t
Margaret [11]

Answer:

When the pressure and the temperature are increased the volume is 285.7 ml.

Explanation:

We can find the new volume by using the Ideal Gas Law:

PV = nRT

Where:

P: is the pressure

V: is the volume

n: is the number of moles

R: is the gas constant

T: is the temperature  

Initially, when V₁ = 200 ml, P₁ = 500 torr and T₁ = 10 °C, we have:

nR = \frac{P_{1}V_{1}}{T_{1}}   (1)  

And finally, when P₂ = 700 torr and T₂ = 20 °C, we have:

nR = \frac{P_{2}V_{2}}{T_{2}}   (2)

By equating (1) with (2):

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}                

V_{2} = \frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}} = \frac{500 torr*200 ml*20 ^{\circ} C}{10 ^{\circ} C*700 torr} = 285.7 ml

Therefore, when the pressure and the temperature are increased the volume is 285.7 ml.                                                

I hope it helps you!                  

3 0
3 years ago
g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

a = \frac{GM}{r^2}

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

F_W = m (\frac{GM}{r^2} ) = 600N

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

dF_W = mda

dF_W = m(-2\frac{GM}{r^3}dr)

dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})

dF_W = -0.3N

Therefore there is a weight loss of 0.3N every kilometer.

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A car has a mass of 1520 kg. While traveling at 20 m⁄s, the driver applies the brakes to stop the car on a wet surface with a 0.
docker41 [41]

Answer:

(a)d₁ = 51.02 m

(b)d₂ =51.02m

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Known data

m=1520 kg  : mass of the  car

μk= 0.4 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the car

We define the x-axis in the direction parallel to the movement of the  car and the y-axis in the direction perpendicular to it.

W: Weight of the block : In vertical direction  downward

FN : Normal force :  In vertical direction  upward

f : Friction force:  In horizontal direction  

Calculated of the W

W= m*g

W=  1520 kg* 9.8 m/s² = 14896 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN - Wy = 0

FN = Wy

FN = 14896 N

Calculated of the f

f = μk* N= (0.4)* (14896 N )

f = 5958.4 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

- f = m*a

-5958.4 = (1520)*a

a  =  (-5958.4) /  ((1520)

a = -3.92 m/s²

(a) displacement of the car (d₁)

Because the car moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d₁ Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

Data:

v₀ = 20 m⁄s

vf = 0

a = --3.92 m/s²

We replace data in the formula (2)  to calculate the distance along the ramp the block reaches before stopping (d₁)

vf²=v₀²+2*a*d ₁

0 = (20)²+2*(-3.92)*d ₁

2*(3.92)*d₁  = (20)²

d₁ = (20)² / (7.84)

d₁  = 51.02 m

(b)  Different car

m₂ = 1.5 *1520 kg

μk₂= 0.4

W₂= m*g

W₂=   (1.5) *1520 kg* 9.8 m/s² = (1.5)*14896 N  

FN₂=  (1.5)*14896 N  

f= 0.4* (1.5)*14896 N  

a = - f/m₂ = - 0.4* (1.5)*14896 N  /(1.5) *1520

a = -3.92   m/s²

vf²=v₀²+2*a*d₂

vf=0 , v₀=20 m⁄s , a = -3.92   m/s²

d₂ = d₁ = 51.02m

6 0
4 years ago
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