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LenaWriter [7]
3 years ago
15

F.r.e.e points :]]] ​

Physics
2 answers:
Hunter-Best [27]3 years ago
6 0

Answer:

lets goooo thanks jdjdjdjdjdjdjdkskdjdjdkdk

Colt1911 [192]3 years ago
4 0
Ok? I don’t know what you want me to do though
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Jayden was given a marshmallow and a syringe in class to experiment with. She placed the marshmallow in the syringe and sealed t
Tju [1.3M]
<h2>Answer:</h2>

The correct option is A.

A) The increased pressure, pushed the molecules closer together, and caused the marshmallow to shrink.

<h2>Explanation:</h2>

Jayden experimented, she placed the marshmallow in the syringe and sealed the end. When she depressed the plunger of the syringe, the pressure increased and pushed the molecules closer together and causes the marshmallow to shrink.

<h2 />
3 0
3 years ago
The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo
a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s

The tire angular velocity before stopping is:

\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s

Also its angular decceleration:

\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2

Using the following equation motion we can findout the angle it makes during the deceleration:

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where \omega = 0 m/s is the final angular velocity of the car when it stops, \omega_0 = 114rad/s is the initial angular velocity of the car \alpha = 14.75 rad/s2 is the deceleration of the can, and \Delta \theta is the angular distance traveled, which we care looking for:

-114^2 = 2*(-14.75)*\Delta \theta

\Delta \theta = 440rad or 440/2π = 70 revelutions

4 0
3 years ago
A student hears a police siren. What would change the frequency that the student hears? Check all that apply.
ser-zykov [4K]
<span>A student hears a police siren.

The arithmetic of the Doppler Effect shows that if the distance between
the source and observer is changing, then the observer hears a different
frequency compared to the frequency actually radiating from the source. 

Thus the first four choices would cause the student to hear a different
frequency:

-- if the student walked toward the police car
-- if the student walked away from the police car
-- if the police car moved toward the student
-- if the police car moved away from the student

The last two choices wouldn't affect the frequency heard by the student,
since the perceived frequency of a sound doesn't depend on its intensity.

-- if the intensity of the siren increased
-- if the intensity of the siren decreased.</span>
4 0
3 years ago
Read 2 more answers
A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and
vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

U=mgh

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

h=4-4cos(30\°)=0.53m

Then, the potential energy is:

U=(55kg)(9.8m/s^2)(0.53m)=285.67J

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

U'=(m_1+m_2)gh'=285.67\ J

From the previous equation you can get h':

h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0
3 years ago
A white billiard ball with mass mw = 1.37 kg is moving directly to the right with a speed of v = 3.14 m/s and collides elastical
Ira Lisetskai [31]

That latest value for the Angle is in Grads, not in Kilograms.

Apply law of conservation of momentum along vertical direction.

m_1v_1sin\theta_1-m_2v_2sin\theta_2=0

v_2=\frac{v_1sin\theta_1}{sin\theta_2}=\frac{sin54}{sin36}v_1 = 1.376v_1

Apply law of conservation of momentum along the horizontal direction

m_1u_1=m_1v_1cos\theta_1+m_2v_2cos\theta_2

u_1=v_1(cos\theta_1+1.376cos\theta_2)

u_1=v_1(cos(54)+1.376cos(36))

u_1=(1.7) v_1

v_1= \frac{3.14}{1.7}=1.84m/s

The second ball velocity is v_2 = (1.376)(1.84)=2.531m/s

The magnitud of final total momentum is

m(v_1+v_2)=(1.37)(2.531+1.84)=5.98kgm/s

The magnitude of final energy is

\frac{1}{2}m(v^2_1+v^2_2)=\frac{1}{2}(1.37)(2.531^2+1.84^2)=6.07J

6 0
3 years ago
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