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2 years ago

14

A squash ball with an initial velocity of 19.21 m/s [W] is hit by a squash racket, changing its velocity to 40.22 m/s [E] in 0.3

12 s. What is the squash ball’s average acceleration?

1 answer:

alexgriva [62]2 years ago

4 0

Acceleration = (change in velocity) / (time for the change)

Change in velocity = 59.43 m/s East

Time for the change = 0.312 s

Acceleration = 190.5 m/s^2 East

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Doe anyone get this

8_murik_8 [283]

Answer:

we know a = F/ M

Explanation:

2 m/s²
0.19 m/s²
9.25 m/ s²
0.04 m/s²
100.39 m/s²

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2 years ago

When driving at night, only use your high-beam headlights___

zzz [600]

Answer:

B

Explanation:

While answer C may sound correct, Answer B is makes more sense. We know you cant use High-beam lights when u cant see ongoing traffic because it could affect the other driver coming across from you. Its good to use it when legal and safe, but in that term I still don't believe there's no reason for HIGH-beamed. That's this leaves B, when you are on u lighted streets.

6 0

2 years ago

A Ford Mustang weighs about 3500 pounds, and can accelerate from 0-60 MPH in about 5 seconds. What force is responsible for this

Hunter-Best [27]

We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

$m = 3500lb = 1587.57kg$

The values of the velocities are,

$\text{Initial Velocity} = V_i = 0$

$\text{Final Velocity} = V_f = 60mph = 26.822m/s$

We know that the acceleration is equivalent to the change of the speed in a certain time therefore

$a = \frac{v_f-v_i}{t}$

$a = \frac{26.822-0}{5}$

$a = 5.36m/s^2$

Now applying the Newton's second law we have,

$F= ma$

$F = (1587.57)(5.36)$

$F = 8516.36N$

Therefore the approximate magnitude is 8516.36N

5 0

3 years ago

German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity (

timama [110]

According to the information given, the Heisenberg uncertainty principle would be given by the relationship

$\Delta x \Delta v \geq \frac{h}{4\pi m}$

Here,

h = Planck's constant

$\Delta v$ = Uncertainty in velocity of object

$\Delta x$ = Uncertainty in position of object

m = Mass of object

Rearranging to find the position

$\Delta x \geq \frac{h}{4\pi m\Delta v}$

Replacing with our values we have,

$\Delta x \geq \frac{6.625*10^{-34}m^2\cdot kg/s}{4\pi (9.1*10^{-31}kg)(0.01*10^6m/s)}$

$\Delta x \geq 5.79*10^{-9}m$

Therefore the uncertainty in position of electron is $5.79*10^{-9}m$

8 0

2 years ago

A 13 cm long tendon was found to stretch 3.7 mm by a force of 12.1 N . The tendon was approximately round with an average diamet

liq [111]

Answer:

young's modulus $=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2$

Explanation:

We have given length of tendon L= 13 cm =0.13 m

Change in length $\Delta L=3.7mm=3.7\times 10^{-3}m$

Force = 12.1 N

Average diameter d =8.8 mm

So $r=\frac{d}{2}=\frac{8.8}{2}=4.4mm=4.4\times 10^{-3}m$

Area $A=\pi\times (4.4\times 10^{-3})^2=60.79\times 10^{-6}m^2$

Now stress $=\frac{force}{area}=\frac{12.1}{60.79\times 10^{-6}}=0.1990\times 10^6N/m^2$

Strain $=\frac{change\ in\ lenght}{length}=\frac{3.7\times 10^{-3}}{0.13}=28.46\times 10^{-3}$

Now young's modulus $=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2$

8 0

3 years ago