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densk [106]
2 years ago
9

1. A runner has an initial speed of 2 [m/s] and slowly speeds up with a constant acceleration of

Physics
1 answer:
zvonat [6]2 years ago
5 0

Answer:

Final Velocity = 4.9 m/s

Explanation:

We are given;. Initial velocity; u = 2 m/s

Constant Acceleration; a = 0.1 m/s²

Distance; s = 100 m

To find the final velocity(v), we will use one of Newton's equations of motion;

v² = u² + 2as

Plugging in the relevant values to give;

v² = 2² + 2(0.1 × 100)

v² = 4 + 20

v² = 24

v = √24

v = 4.9 m/s

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You don’t need time (1) or the force (4) either so just 3,5
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3 years ago
Actiong-reaction of the bat hits the ball.
nevsk [136]

Answer:

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7 0
2 years ago
A voltage of 75 V is placed across a 150 Ω resistor. What is the current through the resistor?
Naddika [18.5K]

Answer:

0.5 A

Explanation:

Applying,

V = IR.................. Equation 1

Where V = Voltage, I = current, R = Resistance.

make I the subject of the equation

I = V/R............... Equation 2

From the question,

Given: V = 75 V, R = 150 Ω

Substitute these values into equation 2

I = 75/150

I = 0.5 A.

Hence the cuurent through the resistor is 0.5 A

6 0
2 years ago
A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle
Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  \alpha =2.538 \  rad/s^2

Explanation:

From the question we are told that

  The mass of the wheel is  m =  6.9  kg

   The radius is  r =  0.69 \  m

    The radius of gyration is  k_G = 0.4\  m

    The angle is  \theta = 47^o

    The force which the massless bar is subjected to F = 22.5 \  N

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that  point A is the  center of rotation.

  Generally the moment of  inertia about A is mathematically represented as

     I_a =  I_G + M* r^2

Here I_G is the moment of inertia about G with respect to the radius of gyration  which is mathematically represented as

    I_G =  M *  k_G

=>I_a = k_G*  M + M* r^2

=>I_a =0.4 * 6.9  + 6.9 * 0.69^2

=>I_a =6.045 \  kg \cdot m^2

Generally the torque experienced by the wheel  is mathematically represented as

       \tau =  F *  cos (47)

=>     \tau =  22.5 *  cos (47)

=>     \tau =  15.34 \ kg \cdot m^2 \cdot s^{-2}

Generally this torque is also mathematically represented as

     \tau = I_a * \alpha

=>   15.34  =  6.045 * \alpha

=>   \alpha =2.538 \  rad/s^2

4 0
3 years ago
Can anyone please help me​
saveliy_v [14]

Answer:its most likely c

Explanation:

5 0
2 years ago
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