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VMariaS [17]
3 years ago
9

The largest salami in the world, made in Norway, was more than 20 m long. If a hungry mouse ran around the salami’s circumferenc

e with a tangential speed of 0.17 m/s, the centripetal acceleration of the mouse was 0.29 m/s2 . What was the radius of the salami?
Physics
1 answer:
Musya8 [376]3 years ago
8 0

Answer: 0.09 m

Explanation:

Centripetal acceleration in terms of tangential speed is:

a_c=\frac{v^2_t}{r}

where r is the radius.

It is given that,

centripetal acceleration of the mouse, a_c=0.29 m/s^2

tangential speed , v_t=0.17m/s

Radius of salami is:

r =\frac{(0.17m/s)^2}{0.29 m/s^2}=0.09m

Thus, the radius of the salami is 0.09 m.

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List five rules of golf etiquette. Explain the correct stance, grip, orientation, and parts of a swing.
Verizon [17]
- keep you eyes on the ball, after you hit the ball you look up, you have to see the club hit the ball

- your grip it really matters on what your comfortable with but first you put your left hand on the club then put your right hand below your left hand, then put your right pinky on top of your index finger, but you can inter lock your pinky and index finger

- DON'T HIT AT THE BALL HIT THREW THE BALL

- parts of the swing, can be pretty tricky to explain, but I'll try my best to explain it, okay so a golf swing is like a pendulum, you swing right to left. Once you lift the club to the right you still look down at the ball. After swimming to the right your have to hit the ball with your right hand strength, in order the ball to fly.

- the way you do your grip defines if the ball will go straight or not. But also the way you swing towards the ball also defines how far the ball is going to go.
8 0
3 years ago
Read 2 more answers
Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

v_2^2 = v_1^2 + 2gh

Where,

V_i = Velocity in each state

g= Gravity

h = Height

Our values are given as,

A_1 = 2.4*10^{-4} m^2

v_1 = 0.8 m/s

h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}

v_2= 1.67 m/s

Applying the concepts of continuity,

A_1v_1 = A_2v_2

We need to find A_2 then,

A_2= \frac{A_1v_1 }{v_2}

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

A_2= 1.14*10^{-4} m2

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 1.14*10^{-4} m2

8 0
3 years ago
Determine whether the center of mass of the system consisting of the earth and moon lies inside or outside the earth. Assume tha
tester [92]

Answer:

R_cm = 4.66 10⁶ m

Explanation:

The important concept of mass center defined by

         R_cm = 1 / M   ∑  x_i m_i

where M is the total mass, x_i and m_i are the position and masses of each body

Let's apply this expression to our case.

Let's set a reference frame where the axis points from the center of the Earth to the Moon,

       R_cm = 1 / M (m_earth 0 + m_moon d)

the total mass is

      M = m_earth + m_moon

     

the distance from the Earth is zero because all mass can be considered to be at its gravimetric center

let's calculate

      M = 5.98 10²⁴ + 7.35 10²²

      M = 6.0535 10₂⁴24 kg

we substitute

      R_cm = 1 / 6.0535 10²⁴ (0 + 7.35 10²² 3.84 )

      R_cm = 4.66 10⁶ m

4 0
3 years ago
(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r
Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface U_1=\frac{GM_em}{R_e}

Gravitational Potential Energy at height h is U_2=\frac{GM_em}{R_e+h}

Energy required to lift the satellite E_1=U_1-U_2

E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}

Now Energy required to orbit around the earth

E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}

\Delta E=E_1-E_2

\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}

E_1=E_2  (given)

\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0

\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0

h=\frac{R_e}{2}

h=3.19\times 10^6\ m

(b)For greater height E_1  is greater than E_2

thus energy to lift the satellite is more than orbiting around earth

4 0
3 years ago
Which statement best describes the skier?
Olegator [25]

its c!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

8 0
3 years ago
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