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hammer [34]
3 years ago
7

A block of ice with mass mA rests on a wedge of ice with mass mB=7.93 kg that is cut so that the surface of the wedge makes an a

ngle of 37.5 with the ground. There is no friction between the pieces of ice or between the ice and the ground. At time t=0 the blocks are released from rest. What is the acceleration of mA in the direction parallel to the ground.
Physics
1 answer:
Alika [10]3 years ago
3 0

Answer:

The answer is 7.52 m/s².

Explanation:

To determine acceleration, N = Ma×g ÷ cos Ф

N sin Ф = ma

a = g Tan Ф m/s²

g = 9.8 m/s² and Ф = 37.5° ( angle for the velocity and gravitational force)

so, a = 9.8 × tan (37.5°)

Acceleration has to be determined in order to get the final answer.

a = 7.5198 m/s²

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A rock is thrown horizontally from a bridge with a speed of 29.0 m/s. if the rock is 23.7 meters above the river at the moment o
mash [69]
It would be 1.5 meters im sure form that distance to me is that nswe

7 0
3 years ago
Give 1 real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is c
OverLord2011 [107]

Answer:

On real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is constant is when making a serve with a lawn tennis racket

How It is an example of impulse is that when a serve is made by moving the bat slowly, the lawn tennis player uses less force and the ball is in contact with the string for longer a period

When however, the lawn tennis player moves the racket faster, with the strings of the racket highly tensioned  he uses more force and the ball also spends less time on the racket to produce the same momentum

Explanation:

The impulse of a force, ΔP is given by the following formula;

ΔP = F × Δt

Where ΔP is constant, we have;

F ∝ 1/Δt

Therefore, for the same impulse, when the force is increased, the time of contact is decreases and vice versa.

7 0
3 years ago
A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

8 0
3 years ago
What best describes why warm ocean currents usually surface currents
hodyreva [135]
Surface currents are caused by winds. They move warm water away from the equator and cool water away from the poles.

Hope it help
3 0
3 years ago
It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f
Elena-2011 [213]

Answer:

B) the change in momentum.

Explanation:

The impulse is defined as the product between the force applied on an object (F) and the duration of the collision (\Delta t):

J=F \Delta t (1)

We can rewrite the force by using Newton's second law, as the product between mass (m) and acceleration (a):

F=ma

So, (1) becomes

J=ma \Delta t

Now we can also rewrite the acceleration as ratio between the change in velocity and change in time: a=\frac{\Delta v}{\Delta t}. If we substitute into the previous equation, we find

J=m\frac{\Delta v}{\Delta t}\Delta t=m\Delta v

And the quantity m\Delta v is equivalent to the change in momentum, \Delta p.

6 0
3 years ago
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