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Valentin [98]
3 years ago
10

A balloon is filled with 1500 mL of air at a pressure of 5 atm. What is the pressure of the balloon if the volume is changed to

2500 mL?
Physics
1 answer:
Eva8 [605]3 years ago
5 0

Answer:

xcfdbrtbfcgctr dsvrd

Explanation:

ds vfg g cd fg

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A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
disa [49]
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, \mu m g. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
ma=-\mu m g
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
2aS=v_f^2-v_i^2
where v_f=0 is the final speed of the skier and v_i=4.75 m/s is the initial speed. Substituting numbers, we find:
S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m
5 0
3 years ago
When finding the upper bound of the density, you put what number in the denominator?
lilavasa [31]
Answer: The result of "the upper bound of the density"  does not go on the denominator. 
So simplified, no. The answer is no.
3 0
3 years ago
How much power does a 2000 kg car need to accelerate from 20 m/s to 35 m/s in 7 seconds?
Alexus [3.1K]

firstly you get your acceleration with the formula, a=v-u/t. Then you use the formula for kinetic energy 1/2mv^2

then you can finally get the answer for power by dividing your previous answer by the time

3 0
3 years ago
A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90
Wewaii [24]

Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

radial acceleration a_r = v^2/r

v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115=  3.175

now,

a= \sqrt{a_t^2+a_r^2}

a= \sqrt{1.47^2+3.175^2}

a= 3.49 m/s^2

3 0
3 years ago
A man ties one end of a strong rope 8.17 m long to the bumper of his truck, 0.524 m from the ground, and the other end to a vert
Kamila [148]

Answer:

2442.5 Nm

Explanation:

Tension, T = 8.57 x 10^2 N

length of rope, l = 8.17 m

y = 0.524 m

h = 2.99 m

According to diagram

Sin θ = (2.99 - 0.524) / 8.17

Sin θ = 0.3018

θ = 17.6°

So, torque about the base of the tree is

Torque = T x Cos θ x 2.99

Torque = 8.57 x 100 x Cos 17.6° x 2.99

Torque = 2442.5 Nm

thus, the torque is 2442.5 Nm.

8 0
3 years ago
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