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12345 [234]
3 years ago
8

Bicycle in its writers have combined a mass of 80 kg in a speed of 6.0 meters per Second what is the magnitude of the average fo

rce needed to bring bicycle in its writers to stop in 4.0 seconds please help !!!
Physics
1 answer:
erik [133]3 years ago
7 0

Answer:

F = 120 N

Explanation:

Force x distance = energy

The bike has energy 1/2 . 80 . 6^2   = 1440 J

You are looking at an example of not reading the question properly.

Impulse = Force . time  = change in momentum

F . 4  = 80 .6

F = 120 N

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Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 re
Luba_88 [7]

Answer:

(a) 152.85 Nm

(b) 1528.5 Nm

Explanation:

According to the formula of power

P = τ ω

ω = 2 π f

(a) f = 2500 rpm = 2500 / 60 = 41.67 rps

So, 40 x 1000 = τ x 2 x 3.14 x 41.67

τ = 152.85 Nm

(b) f = 250 rpm = 250 / 60 = 4.167 rps

So, 40 x 1000 = τ x 2 x 3.14 x 4.167

τ = 1528.5 Nm

3 0
2 years ago
what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?
nignag [31]

Using Newton's second law of motion:

F=ma ;  [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)


Given:                      Find:                   Formula:                  Solve for m:

F: 2500N                 mass:?                F=ma Eq.1              m=F/a  Eq. 2

a= 200m/s^2  


Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

8 0
3 years ago
What's ur fav song?<br><br> Fr33 Pointtsss &lt;3
arsen [322]

Answer:

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Haha Rick rolled you

Explanation:

jk my favourite song is Thunder, Despacito

3 0
2 years ago
Read 2 more answers
Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const
kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is  L=  20cm = 0.2m

Thermal conductivity of the wall is  K = 2.79 W/m·K

Temperature at the left side surface is T₁ =  50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is  h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

Q_{conduction} = Q_{convection}

Expression for the heat conduction process is

Q_{conduction} = \frac{K(T_1 - T)}{L}

Expression for the heat convection process is

Q_{convection} = h(T_2 - T)

Substitute the expressions of conduction and convection in equation above

Q_{conduction} = Q_{convection}

\frac{K(T_1 - T_2)}{L} = h(T_2 - T)

Substitute the values in above equation

\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC

Now heat flux through the wall can be calculated as

q_{flux} = Q_{conduction} \\\\q_{flux}  = \frac{K(T_1 - T_2)}{L}\\\\q_{flux}  = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0
2 years ago
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c
Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe,d_1=0.573 m

v_1=13.5 m/s

v_2=5.83 m/s

Volume flow rate of the petroleum along the pipe=Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})

Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s

By equation of continuity

A_1v_1=A_2v_2

\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2

d^2_2=\frac{v_1}{v_2}d^2_1

d_2=\sqrt{\frac{v_1}{v_2}}d_1

d_2=0.573\sqrt{\frac{13.5}{5.83}}

d_2=0.87 m

d_2=0.87\time 100=87 cm

1 m=100 cm

4 0
3 years ago
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