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kolezko [41]
4 years ago
6

4. Look at the graph above. What information is on the graph and what does it mean? Please help!!!

Physics
1 answer:
Aliun [14]4 years ago
6 0

Answer:

CO2 increases in the winter

Explanation:

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Shearing of the wool is done with special instruments called_____​
Kobotan [32]

Answer:

The machine used is called a squaring shear, power shear, or guillotine.

Explanation:

5 0
3 years ago
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Can y’all please help me with this 3 part question?
klio [65]

Answer:

Vf = 210 [m/s]

Av = 105 [m/s]

y = 2205 [m]

Explanation:

To solve this problem we must use the following formula of kinematics.

v_{f} =v_{o} +g*t

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (released from the rest)

g = gravity acceleration = 10 [m/s²]

t = time = 21 [s]

Vf = 0 + (10*21)

Vf = 210 [m/s]

Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)

The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)

Av = (210 + 0)/2

Av = 105 [m/s]

To calculate the distance we must use the following equation of kinematics

v_{f} ^{2} =v_{o} ^{2} +2*g*y\\\\(210)^{2} = 0 + (2*10*y)

44100 = 20*y

y = 2205 [m]

5 0
3 years ago
If a car goes along a straight road heading east and speeds up from 45 ft/s to 60. ft/s in 5 sec, calculate the
Lapatulllka [165]

Answer:

Acceleration = 0.9144 m/s^2

Explanation:

Initial speed = 45 ft/s

Final speed = 60 ft/s

Time = 5 sec

Acceleration = a = (v-u) / t

                     = 60-45 / 5

                     = 0.9144 m/s^2

8 0
4 years ago
Read 2 more answers
Please help!<br> If an object has a mass of 500 g and floats, what volume of water must it displace?
mafiozo [28]

It must displace at least 500 milliliters (0.5 liter) of water, in order to float in water.

4 0
3 years ago
A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing
Damm [24]

Given:

Shaft Power, P = 7.46 kW = 7460 W

Speed, N = 1200 rpm

Shearing stress of shaft, \tau _{shaft} = 30 MPa

Shearing stress of key, \tau _{key} = 240 MPa

width of key, w = \frac{d}{4}

d is shaft diameter

Solution:

Torque, T = \frac{P}{\omega }

where,

\omega = \frac{2\pi  N}{60}

T = \frac{7460}{\frac{2\pi  (1200 )}{60}} = 59.365 N-m

Now,

\tau _{shaft} = \tau _{max} = \frac{2T}{\pi (\frac{d}{2})^{3}}

30\times 10^{6} = \frac{2\times 59.365}{\pi (\frac{d}{2})^{3}}

d = 0.0216 m

Now,

w =  \frac{d}{4} =  \frac{0.02116}{4} = 5.4 mm

Now, for shear stress in key

\tau _{key} = \frac{F}{wl}

we know that

T = F \times r =  F. \frac{d}{2}

⇒ \tau _{key} = \frac{\frac{T}{\frac{d}{2}}}{wl}

⇒ 240\times 10^{6} = \frac{\frac{59.365}{\frac{0.0216}{2}}}{0.054l}

length of the rectangular key, l = 4.078 mm

7 0
3 years ago
Read 2 more answers
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