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3 years ago

How are mm hg torr psi atm and pa related to each other mathematically?

1 answer:

7 0

Mathematically the relationship is;
1 atm is equivalent to 760 mmHg
1 atm is equivalent to 760 torr
Therefore; 1 torr ≈ 1 mmHg
1 atm is equivalent to 101,325 Pa
1 atm is equivalent to 14.69594 psi

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A 250 kg car has 6875 kg•m/s of momentum. What is it’s velocity?

liraira [26]

Answer:

v = 27.5 m/s

Explanation:

p = m × v

6,875 = 250 × v

250v = 6,875

v = 6,875/250

v = 27.5 m/s

5 0

2 years ago

How does the number 10 relate to light, sound, force and motion?

Oksana_A [137]

Answer:

number 10 is the value of acceleration due to gravity.

8 0

3 years ago

Determine the amount of work done by the engine of a car with a mass of 2500 kg when it accelerates from 45 mph to 65 mph. (Use

Y_Kistochka [10]

Answer:

amount of work done, W = 549.36 kJ

Given:

mass of a car engine, m = 2500 kg

initial velocity, u = 45 mph

final velocity, v = 65 mph

1 mile = 1609

Solution:

We know that 1 hour = 3600 s

Now, velocities in m/s are given as:

u = 45 mph = $\frac{45\times 1609}{3600}$ = 20.11 m/s

v = 65 mph = $\frac{65\times 1609}{3600}$ = 29.05 m/s

Now, the amount of work done, W is given by the change in kinetic energy of the car and is given by:

W = $\frac{1}{2}m\Delta v^{2}$

W = $\frac{1}{2}m\times (v^{2} - u^{2})$

W = $\frac{1}{2}2500\times (29.05^{2} - 20.11^{2})$

W = 549.36 kJ

3 0

3 years ago

slader A transcontinental flight of 4890 km is scheduled to take 40 min longer westward than eastward. The airspeed of the airpl

11111nata11111 [884]

Answer:

$v_{s}=65.2km/h$

Explanation:

Given data

Flight distance S=4890 km

Time difference Δt=t₂-t₁=40 min

Air speed of plane=980 km/h

To find

Speed of jet stream

Solution

When moving in the same direction as the jet stream time taken as t₁=d/(v+vs),v is velocity of plane and vs is velocity of plane

While moving in opposite direction t₂=d/(v+vs)

$t_{2}-t_{1}=\frac{d}{(v-v_{s}) } - \frac{d}{(v+v_{s}) }\\t_{2}-t_{1}=\frac{d(v+v_{s})-d(v-v_{s})}{(v-v_{s})(v+v_{s})} \\t_{2}-t_{1}=\frac{2dv_{s}}{(v)^{2} -(v_{s})^{2} }\\0.666667h=\frac{2(4890km)v_{s}}{(980km/h)^{2} -(v_{s})^{2} }\\0.666667((980km/h)^{2} -(v_{s})^{2})=9780v_{s}\\640267-0.666667(v_{s})^{2}-9780v_{s}=0\\0.666667(v_{s})^{2}+9780v_{s}-640267=0$

Apply quadratic formula to solve for vs

$v_{s}=65.2km/h$

4 0

3 years ago

Learning Goal: To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following gene

kaheart [24]

Answer:

so the transverse displacement is 0.089963 m

Explanation:

Given data

equation y(x,t)=Acos(kx−ωt)

speed v = 9.00 m/s

amplitude A = 9.00 × 10^−2 m

wavelength λ = 0.480 m

to find out

the transverse displacement

solution

we know

v = angular frequency / wave number

and

wave number = 2 $\pi$ / λ = 2 $\pi$ / 0.480 = 13.0899 $m^{-2}$

angular frequency = v k

angular frequency = 9.00 × 13.0899

angular frequency = 117.8097 rad/sec = 118 rad/sec

equation y(x,t)=Acos(kx−ωt)

y(x,t)=9.00 × 10^−2 cos(13.0899 x−118t)

when x =0 and and t = 0

maximum y(x,t)= 9.00 × 10^−2 cos(13.0899 (0) − 118 (0))

maximum y(x,t)= 9.00 × 10^−2 m

and when x = x = 1.59 m and t = 0.150 s

y(x,t)=9.00 × 10^−2 cos(13.0899 (1.59) −118(0.150) )

y(x,t)=9.00 × 10^−2 × (0.99959)

y(x,t) = 0.089963 m

so the transverse displacement is 0.089963 m

5 0

3 years ago