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4 years ago

If the skater coasts only around 65 degrees of the circle, find the magnitude of his displacement vector.

1 answer:

4 0

We are given a skater who coasts only around 65 degrees of the circle. The magnitude of the vector is determined by converting 65 degrees to radians:

65//180 = 0.36 radians

The displacement of the vector is 65 degrees east.

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An object moves with a speed of 30 km/h for 15 minutes and then with a speed of 60km/h for the next 15 minutes. Find the total d

Finger [1]

Answer:

Total distance = 22.5 kilometers

Explanation:

Let the first distance be A.

Let the second distance be B

Given the following data;

Speed A = 30km/h

Time A = 15 minutes to hours = 15/60 = 0.25 hour

Speed B = 60km/h

Time B = 15 minutes to hours = 15/60 = 0.25 hour

Now, to find the respective distance traveled;

Distance A = speed A * time A

Distance A = 30 * 0.25

Distance A = 7.5 km

Distance B = speed B * time B

Distance B = 60 * 0.25

Distance B = 15 km

Next, we calculate the total distance traveled;

Total distance = Distance A + distance B

Total distance = 7.5 + 15

Total distance = 22.5 kilometers.

Therefore, the total distance covered by the object is 22.5 kilometers.

5 0

3 years ago

1. Is the relationship between velocity and centripetal force a direct, linear relationship or is it a nonlinear square relation

Svetllana [295]

Answer:

non linear square relationship

Explanation:

formula for centripetal force is given as

a = mv^2/r

here a ic centripetal acceleration , m is mass of body moving in circle of radius r and v is velocity of body . If m ,and r are constant we have

a = constant × v^2

a α v^2

hence non linear square relationship

7 0

3 years ago

Un auto si muove lungo una strada rettilinea

Valentin [98]

Answer:

Un'auto si muove lungo un percorso rettilineo con velocità variabile come mostrato in figura. Quando l'auto è in possesso di A, la sua velocità è 10 ms-1 e quando è in posizione B, la sua velocità è 20 ms-1. Se l'auto impiega 5 secondi per spostarsi da A a B, trova l'accelerazione dell'auto.

Explanation:

8 0

3 years ago

A hoop and a solid disc are relased from rest

Murljashka [212]

Answer:

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

For the hoop:

$K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0$

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

$I_h = m_hR^2$

Let's continue with the energy equations:

$m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}$

Similarly for the solid disk with a moment of inertia of (1/2)mR^2:

$K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}$

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

$v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2$

where a is the linear acceleration.

The relation with the angular and linear acceleration is

$a = \alpha R$

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

$\alpha = 2.235/R ~m/s^2$

4 0

4 years ago

Read 2 more answers

In which type of Eclipse do u see the Sun in the form of a ring

Helga [31]

The type of eclipse in which one sees the sun in the form of a ring is ANNULAR SOLAR ECLIPSE.
Solar eclipse occurs when the moon casts a shadow on the earth.
There are three basic types of solar eclipse, these are total, partial and annular solar eclipse. The annular solar eclipse occurs when the moon covers the sun's center thereby leaving the sun visible outer edges to form a ring of fire around the moon.

8 0

3 years ago