If you push a book a against a wall hard enough , it will not slide down even through gravity is pulling it

Superman is flying 54.5 m/s when he sees

Nady [450]

348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations $d=v_{0} t+\frac{at^{2} }{2}$ and $v_{f} =v_{0} +at$ where $d$ is distance, $v_{0}$ is the initial velocity, $v_{f}$ is the final velocity, $t$ is time, and $a$ is aceleration.

Superman's initial velocity is $v_{0}=54.5\frac{m}{s}$ , and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know $d$ , $v_{0}$ and $t$ , we have to find the aceleration $a$ in order to find $v_{f}$ .

From the equation $d=v_{0} t+\frac{at^{2} }{2}$ we have to clear $a$ , getting the equation as follows: $a=\frac{2(d-v_{0}t) }{t^{2} }$ .

Substituting the values:

$a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}$

To find $v_{f}$ we use the equation $v_{f} =v_{0} +at$ .

Substituting the values:

$v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}$

5 0

3 years ago

A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a

aleksley [76]

Answer:

T₂ =602 °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

$\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\$

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602 °C

5 0

3 years ago

A stone that starts at rest free falls for 7.0 s. How far does the stone fall in this time?

vredina [299]

Gravitational acceleration is approx 9.8 m/s
Time is 7s

a=9.8 m/s
t=7s

a = d/t^2

therefore:

d = a * t^2

d = 9.8 * 7^2

d = 9.8 * 49

d = 480.2 [m]

7 0

3 years ago

How much work is done? A Net Force of 9.0 N acts through a distance of 3.0 m in a time of 3.0 s. The answers are 3.0 J, 9.0 J, 2

Vadim26 [7]

If the force and the motion are along the same direction (like it is here) then work is force*distance. The time doesn't come into play until you want the power used. So here
W=9.0*3.0=27J

5 0

3 years ago

An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal

borishaifa [10]

Answer and Explanation:

If the ant was to crawl 50cm to the right, then come back 30 cm, then the total distance walked would be 80cm.

- Combine 50cm and 30cm to get 80 cm.

For displacement, the answer is 20 cm.

- When calculating displacement, you use the initial (starting) distance. and subtract that from the final distance, giving you the displacement, or the amount traveled from the starting point to the final point if you were to make a straight line from the starting point to end point. (0 to 50, then back 30 the same direction, so subtract 30 from 50 to get 20)

#teamtrees #PAW (Plant And Water)

I hope this helps!

8 0

2 years ago

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