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3 years ago

Studies show that the amount of heat stored in the ocean is increasing. What effect might this have?

Physics

2 answers:

lianna [129]3 years ago

8 0

Answer:

Ocean currents might change course

Explanation:

g100num [7]3 years ago

4 0

This warming will have an effect on the bicarbonate buffer of the ocean.

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slega [8]

A chemist is likely to:
1. analyze the ingredients in ice cream
2. determine how to separate gasoline from other substances in petroleum

7 0

3 years ago

Read 2 more answers

What kind of charge is produced when a silk shirt is rubbed with human body

jeka94

Answer:

Frictional charges

Explanation:

6 0

3 years ago

Question 1 (1 point)

deff fn [24]

Answer:

Travelled 18 km, they are 6 km from home.

Explanation:

12/2 (halfway) is 6km. So, 6 + 12 would be 18 km, total amount travelled. The total distance of the trip would be 24 km (12 km out, 12km back) if they travelled 12+6 (18km) then they only have 6 km more to go.

5 0

3 years ago

On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m

Nastasia [14]

Answer:

$F = 1.489*10^{-7} N$

Explanation: Weight of space probes on earth is given by: $W= m*g$

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 $\frac{m}{s^{2} }$ )

Therefore,

$m_{1} = \frac{14500}{9.81}$

$m_{1} = 1478.08 kg$

Similarly,

$m_{2} = \frac{4800}{9.81}$

$m_{2} = 489.29 kg$

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

$F = \frac{Gm_{1} m_{2}}{R^{2} }$

G= gravitational constant ( $6.67 * 10^{-11} m^{3} kg^{-1} s^{-2}$ )

$m_{1} , m_{2}$ = masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

$F = 1.489*10^{-7} N$

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

7 0

3 years ago

A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is

Charra [1.4K]

Answer:

M is equal to m

Explanation:

In case we say that the green block's mass m is less than red block's mass M, then the green block would have bounced and moved back to the left instead of coming to rest. The other case where if mass of green block's mass m would have been greater than the red block's mass M, the green block would have kept moving to the right instead of coming to rest. After collision, the red block moves to the right because of exchange of velocities. Therefore, m=M since m comes to rest and M moves to the right

In any collision, as it is asumed that no external forces can act during the collision, momentum must be conserved.

So, if we call p₁ to the momentum before collision, and p₂ to momentum after it, taking into account the information above, we can write the following:

p₁ = mv₁ + M.0 = p₂ = m.0 + Mv₂ ⇒ mv₁ = Mv₂

From the question, we also know that it was an elastic collision.

In elastic collision, added to the momentum conservation, it must be conserved the kinetic energy also.

So, if we call k₁ to the kinetic energy prior the collision, and k₂ to the one after it, we can write the following:

k₁ = 1/2 m(v₁)² + 1/2 M.0 = k₂ = 1/2m.0 + 1/2M(v₂)² ⇒ m(v₁)² = M(v₂)²

Mathematically, the only way in which both equations be true, should be with v₁ = v₂, which is only possible if m=M too.

In this type of collision, it is said that the energy transfers from one mass to the other.

8 0

3 years ago