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Snezhnost [94]
4 years ago
14

MathPhys Help pls Tysm

Physics
2 answers:
andrew-mc [135]4 years ago
7 0

Answer:

8.75

Explanation:

HACTEHA [7]4 years ago
3 0

Answer:

8.75

Explanation:

First, find the force of friction.

Kinetic energy = work done by friction

½ mv² = Fd

½ (3.9 kg) (2.9 m/s)² = F (1.4 m)

F = 11.7 N

Next, find the distance at the new velocity.

Kinetic energy = work done by friction

½ mv² = Fd

½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d

d = 8.75 m

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Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur
lidiya [134]

Answer:

22.22m/s

Explanation:

The momentum before a collision = momentum after collision so...

work out the momentum of the first object (the bullet)

its p = mv

0.04 kg × 300 m/s  = 0.54 kg × v

rearrange this to find v which is 0.04 x 300 = 12

so 12 = 0.54 x v

    12/0.5 = v

    v = 22.22m/s

hope this helps!

7 0
3 years ago
Electrons in excited hydrogen atoms are in the
Sidana [21]
I can see three different transitions here:

3 --> 1

3 --> 2
followed by
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4 0
3 years ago
What is the most likely elevation of point B?<br>a. 150 ft<br>b. 200 ft<br>c. 125 ft<br>d. 225 ft​
zepelin [54]

Answer:

D

Explanation:

erm i think its 225 ft?

7 0
4 years ago
Read 2 more answers
A water strider bug is supported on the surface of a pond by surface tension acting along the interface between the water and th
Alik [6]

Answer:

minimum interface length = 1.36 mm

Explanation:

given data

weight of the bug = 10^{-4} N

solution

we will apply here Surface Tension formula that is

Surface Tension ,σ = Force ÷ length      ........................1

and we consider here surface tension for water is 7.34 × 10^{-2} N/m

so that here minimum interface length needed to support the bug is

minimum interface length = Force ÷ σ

minimum interface length = 10^{-4}  ÷ 7.34 × 10^{-2}

minimum interface length = 1.36 mm

                                                                                     

                                                                                     

4 0
3 years ago
The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared ​(units are in​ meters). If
ivanzaharov [21]

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

y=3+0.8x-0.4x^2........(1)

The x component of constant velocity, v_x=5\ m/s

We need to find the resultant velocity at the point (2,3).

Let \dfrac{dx}{dt}=v_x and \dfrac{dy}{dt}=v_y

Differentiating equation (1) wrt t as,

\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}

v_y=0.8\times v_x-0.8x\times v_x

v_y=0.8v_x(1-x)

When x = 2 and v_x=5\ m/s

So,

v_y=0.8\times 5\times (1-2)

v_y=-4\ m/s

Resultant velocity, v=\sqrt{v_x^2+v_y^2}

v=\sqrt{5^2+(-4)^2}

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

8 0
3 years ago
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