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Stolb23 [73]
3 years ago
6

A ball is initially moving with a velocity 0.5m/s. Its velocity decreases at a rate of 0.05m/s^2. How much time will it take to

stop? How much distance will the ball travel before it stops
Physics
1 answer:
gogolik [260]3 years ago
5 0

Answer:  t = 10 s , s = 2,5 m

Explanation:   (1) Speed v = v0 + at and (2) distance s = v0t + ½at².

Now  v0 = 0.5 m/s , v = 0 and a = -0.05 m/s²

Solve from equation (1)   t = -v0 / a = - 0.5 m/s / -0.05 m/s² =   10 s

Add t = 10 s to equation (2 )   s = 0.5 m/s · 10 s + 0.5· (-0.05 m/s²)· (10 s)²

= 5 m - 2,5 m = 2,5 m

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A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o
blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

P=\rho \times g \times h

Here

  • P is the pressure which is given as 68 kPa.
  • ρ is the density of the oil whose SG is 0.92. It is calculated as

                                       \rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\

  • g is the gravitational constant whose value is 9.8 m/s^2
  • h is the height which is to be calculated

                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

So the height of column is 7.54m

a-3

By the relation of volume and density

M=\rho \times V

Here

  • ρ is the density of the oil which is 920 kg/m^3
  • V is the volume of cylinder with diameter 16m calculated as follows

                             V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3

Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

                            \Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\

Now corrected pressure is as

P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

4 0
3 years ago
Speed has ____ magnitude and _____ direction ??
ArbitrLikvidat [17]
Speed has only magnitude and no direction. 
6 0
3 years ago
A 400.0 kg storage box is held 10 m above ground by a forklift. What is its gravitational potential energy? (PE = mgh)
castortr0y [4]

Answer: D.) 39,200 J

Via the equation of potential energy PE = mgh where m is mass, g is the average gravity on earth and h is the height.  In this case m = 400 kg, g = 9.8, h = 10 m thus:

P.E.=(400kg)(9.8\frac{m}{s^2} )(10m)=39,200 J

P.E.= 39,200 Joules

7 0
3 years ago
Read 2 more answers
Which two options would INCREASE the electric force between two charged particles?
atroni [7]

The electric force between two charged particles can be increased by decreasing the distance between the two particles.

<h3>How to increase electric force between two charged particles.</h3>

The technique of decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. while

increasing the separation distance between objects decreases the force of attraction or repulsion between the objects.

Read more on Electric Force:

brainly.com/question/17692887

#SPJ1

7 0
2 years ago
A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.
wlad13 [49]

Answer:

12m/s

Explanation:

v_f=v_o+at

Let's call the velocity that the car maintains for 10 seconds v_f_1, and the final velocity v_f_2.

v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s

Hope this helps!

5 0
3 years ago
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