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Lera25 [3.4K]
2 years ago
8

Convection currents in air and water occur because _

Physics
1 answer:
Diano4ka-milaya [45]2 years ago
7 0

warm fluids are less dense than cold fluids

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Carter pushes a bag full of basketball jerseys across the gym floor. The he pushes with a constant force of 21 newtons. If he pu
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Carter needs a power equal to 63 W to be able to push the bag full of Jersey. This is by using the formula: Power is equal to the product of Force applied and Displacement all over time traveled.
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Which statement best explains the movement of electric current from the clouds to the ground during a lightning storm? .
olga2289 [7]
"The <span>ground is positively charged and the clouds are negatively charged " is the statement among the statements given in the question that </span><span>best explains the movement of electric current from the clouds to the ground during a lightning storm. The correct option among all the options that are given in the question is the third option or option "C". </span>
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2 years ago
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Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of be
ANTONII [103]

Answer:

W=315 x 10⁵ J

Explanation:

Given that

F= 2.5 x 10⁵ N

d= 90 m

K.E.=5.4 x 10⁷ J

We know that work done by all force is equal to the change in kinetic energy

Lets take work done by catapult is W

W + F.d= K.E.

W= 5.4 x 10⁷ -  2.5 x 10⁵  x 90 J

W= (540 - 25 x 9) 10⁵ J

W=315 x 10⁵ J

5 0
3 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

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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