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dusya [7]
3 years ago
8

For you, do exercises boost your confidence? how?​

Physics
2 answers:
mihalych1998 [28]3 years ago
8 0

Answer:

yes

Explanation:

Because it helps you understand what kind of question will be coming in the exam

SVETLANKA909090 [29]3 years ago
7 0

Answer:

yes for me

Explanation:

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The tape in a videotape cassette has a total
Softa [21]

Answer:

1.37 rad/s

Explanation:

Given:

Total length of the tape is, d= 297 m

Total time of run is, t = 2.1 hours

We know, 1 hour = 3600 s

So, 2.1 hours = 2.1 × 3600 = 7560 s

So, total time of run is, t= 7560 s

Inner radius is, r = 10\ mm = 0.01\ m


Outer radius is, R = 47\ mm = 0.047\ m


Now, linear speed of the tape is, v =\frac{d}{t}=\frac{297}{7560}=0.039\ m/s


Let the same angular speed be \omega.

Now, average radius of the reel is given as the sum of the two radii divided by 2.

So, average radius is, R_{avg}=\frac{R+r}{2}=\frac{0.047+0.01}{2}=\frac{0.057}{2}=0.0285\ m


Now, common angular speed is given as the ratio of linear speed and average radius of the tape. So,

\omega=\dfrac{v}{R_{avg}}\\\\\\\omega=\dfrac{0.039}{0.0285}\\\\\\\omega=1.37\ rad/s


Therefore, the common angular speed of the reels is 1.37 rad/s.

5 0
3 years ago
Mickey walks Pluto 4 miles North to the dog park. Mickey finds beautiful
yanalaym [24]
Answer is 12 milesBecause yeah
3 0
3 years ago
If a car travels 400m in 20 seconds how fast is it going? 20 m/s
scoundrel [369]

Answer:

20m/s

Explanation:

Speed = distance / time

Speed = 400/20

Speed = 20

5 0
3 years ago
A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
3 years ago
A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a
Softa [21]

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

F_N = F_C - F_G = m\frac{v^{2} }{r} - mg

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

E_{tot}=mgh + \frac{1}{2} mv^{2}

At the top of the hill:

E_{tot}= mgh_{hill}

At the top of the loop:

E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}

Setting both energies equal and canceling the mass m gives:

gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}

Solving for v:

v^{2} = 2g(h_{hill}-h_{loo}_p)

Using v in the first equation:

F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg

F_N = 955.5N

7 0
3 years ago
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