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jasenka [17]
3 years ago
7

A roller coaster accelerates from 0 m/s to 25 m/s in 5 seconds. Determine the acceleration of the roller coaster.

Physics
1 answer:
pashok25 [27]3 years ago
5 0
Acceleration = change of velocity / time taken = 25/5 = 5 m/s/s
You might be interested in
A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the
bazaltina [42]

Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

diameter = 2.0m = 2\times 10^{-3}

Extended by = 0.25mm = 0.25\times 10^{-3}

Based on the above information, the calculation is as follows

a. The Stress of the wire is

= \frac{force\ applied}{area\ of \ circle}

here area of circle = perpendicular to the are i.e cross-sectional  i.e

= \frac{\pi d^{2}}{4}

= \frac{\pi(2\times 10^{-3})^2}{4}

Now place these above values to the above formula

= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}

= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

b. Now the strain of the wire is

= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}

= 5 \times 10^{-5}

3 0
3 years ago
If the pressure of gas is doubled and its absolute temperature is quadrupled, the volume is what factor times the original?
Lapatulllka [165]

Answer:

Volume will increase by factor 2

So option (A) will be correct answer  

Explanation:

Let initially the volume is V pressure is P and temperature is T

According to ideal gas equation PV=nRT, here n is number of moles and R is gas constant

So V=\frac{nRT}{P}....................eqn 1

Now pressure is doubled and temperature is quadrupled

So new volume V_{new}=\frac{nR4T}{2P}=\frac{2nRT}{P} ........eqn 2

Now comparing eqn 1 nad eqn 2

V_{new}=2V

So volume will increase by factor 2

So option (A) will be correct answer

8 0
2 years ago
A ball player catches a ball 3.2 s after throwing it vertically upward. what height did it reach?
cluponka [151]

At the top of the height, the velocity is zero and acceleration is negative of acceleration due to gravity ( i.e  -9.8 m/s^2).

The time of the ball in air is 3.2 s, so ascending time is 3.2/2=1.6 s.

Therefore from kinematic equation,

v = u + gt

Substituting the values we get,

0= u - 9.8 (1.6)\\\\u=15.7 m/s, Here v = 0 at top.

Now from equation,

h=ut+\frac{1}{2} gt^2,  here h is the height .

So,

h=(15.7 m/s) (1.6s)-\frac{1}{2} 9.8m/s^2(1.6)^2\\\\ h=25.12m-12.54m\\\\h=12.48 m.

Thus, the ball reached at its maximum height of 12.48 m.

8 0
3 years ago
A cylinder with a piston contains 0.300 mol of oxygen at 2.50×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The
alukav5142 [94]

Answer:

a) W =  900   J.  b) Q =  3142.8   J . c) ΔU =  2242.8   J. d) W = 0. e) Q =   2244.78   J.  g) Δ U  =  0.

Explanation:

(a) Work done by the gas during the initial expansion:

The work done W for a thermodynamic constant pressure process is given as;

W  =  p Δ V

where  

p  is the pressure and  Δ V  is the change in volume.

Here, Given;

P 1 = i n i t i a l  p r e s s u r e  =  2.5 × 10^ 5   P a

T 1 = i n i t i a l   t e m p e r a t u r e  =  360   K

n = n u m b er   o f   m o l e s  =  0.300  m o l  

The ideal gas equation is given by  

P V = nRT

where ,

p  =  absolute pressure of the gas  

V =  volume of the gas  

n  =  number of moles of the gas  

R  =  universal gas constant  =  8.314   K J / m o l   K

T  =  absolute temperature of the gas  

Now we will Calculate the initial volume of the gas using the above equation as follows;

PV  =  n R T

2.5 × 10 ^5 × V 1  =  0.3 × 8.314 × 360

V1 = 897.91 / 250000

V 1  =  0.0036   m ^3  = 3.6×10^-3 m^3

We are also given that

V 2  =  2× V 1

V2 =  2 × 0.0036

V2 =  0.0072   m^3  

Thus, work done is calculated as;

W  =  p Δ V  = p×(V2 - V1)

W =  ( 2.5 × 10 ^5 ) ×( 0.0072  −  0.0036 )

W =  900   J.

(b) Heat added to the gas during the initial expansion:

For a diatomic gas,

C p  =  7 /2 ×R

Cp =  7 /2 × 8.314

Cp =  29.1  J / mo l K  

For a constant pressure process,  

T 2 /T 1  =  V 2 /V 1

T 2  =  V 2 /V 1 × T 1

T 2  =  2 × T 1  = 2×360

T 2  =  720  K

Heat added (Q) can be calculated as;  

Q  =  n C p Δ T  = nC×(T2 - T1)

Q =  0.3 × 29.1 × ( 720  −  360 )

Q =  3142.8   J .

(c) Internal-energy change of the gas during the initial expansion:

From first law of thermodynamics ;

Q  =  Δ U + W

where ,

Q is the heat added or extracted,

Δ U  is the change in internal energy,

W is the work done on or by the system.

Put the previously calculated values of Q and W in the above formula to calculate  Δ U  as;

Δ U  =  Q  −  W

ΔU =  3142.8  −  900

ΔU =  2242.8   J.

(d) The work done during the final cooling:

The final cooling is a constant volume or isochoric process. There is no change in volume and thus the work done is zero.

(e) Heat added during the final cooling:

The final process is a isochoric process and for this, the first law equation becomes ,

Q  =  Δ U  

The molar specific heat at constant volume is given as;

C v  =  5 /2 ×R

Cv =  5 /2 × 8.314

Cv =  20.785  J / m o l   K

The change in internal energy and thus the heat added can be calculated as;  

Q  = Δ U  =  n C v Δ T

Q =  0.3 × 20.785 × ( 720 - 360 )

Q =   2244.78   J.

(f) Internal-energy change during the final cooling:

Internal-energy change during the final cooling  is equal to the heat added during the final cooling Q  =  Δ U  .

(g) The internal-energy change during the isothermal compression:

For isothermal compression,

Δ U  =  n C v Δ T

As their is no change in temperature for isothermal compression,  

Δ T = 0 ,  then,

Δ U  =  0.

8 0
2 years ago
If a racket strikes a ball (m = 100 g) with a force of 2000 N for 0.3 s., calculate the velocity the ball has as it leaves the r
irina1246 [14]
Force can be calculated by the product of mass and the acceleration of the object in motion. Acceleration is the change of velocity per change in time. The expression would be:

F = ma

where
a = Δv/Δt
The object is from rest and time 0. Acceleration would reduce to:
a = v/t

F = m(v/t)
2000 = 100 (v / 0.3)
v = 6 m/s -------> OPTION 1

3 0
3 years ago
Read 2 more answers
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