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Aleksandr-060686 [28]
3 years ago
12

Coach Hogue is riding his motorcycle in a circle on wet pavement. Suddenly the bike slides out from under him. What failed to pr

ovide enough Centripetal Force to maintain coach Hogue's circular motion Group of answer choices Inertia Force Force of Friction Spring Force Force of Gravity
Physics
1 answer:
Arisa [49]3 years ago
5 0

Answer:

The correct option is;

Force of Friction

Explanation:

As coach Hogue rode his motorcycle round in circle on the wet pavement, the motorcycle and the coach system tends to move in a straight path but due to intervention by the coach they maintain the circular path

The motion equation is

v = ωr and we have the centripetal acceleration given by

α = ω²r and therefore centripetal  force is then

m×α = m × ω²r = m × v²/r

The force required to keep the coach and the motorcycle system in their circular path can be obtained by the impressed force of friction acting towards the center of the circular motion.

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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
What is the wavelength associated with an electron with a velocity of 4.8X10s m/s? (Mass of the electron is 9.1X10-31kg)?
amid [387]

Answer:

1.52 nm

Explanation:

Using the De Broglie wavelength equation,

λ = h/p where λ = wavelength associated with electron, h = Planck's constant = 6.63 × 10⁻³⁴ Js and p = momentum of electron = mv where m = mass of electron = 9.1 × 10⁻³¹ kg and v = velocity of electron = 4.8 × 10⁵ m/s

So, λ = h/p

λ = h/mv

substituting the values of the variables into the equation, we have

λ = h/mv

λ = 6.63 × 10⁻³⁴ Js/(9.1 × 10⁻³¹ kg × 4.8 × 10⁵ m/s)

λ = 6.63 × 10⁻³⁴ Js/(43.68 × 10⁻²⁶ kgm/s)

λ = 0.1518 × 10⁻⁸ m

λ = 1.518 × 10⁻⁹ m

λ = 1.518 nm

λ ≅ 1.52 nm

4 0
3 years ago
f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t
inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

h(t)=35t-0.83t^2

We know that the rate of change of displacement is equal to the velocity of an object.

v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t

Velocity of the arrow after 3 seconds will be :

v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0
3 years ago
Is it more fun on a swing with low or high amplitude? Explain.
marusya05 [52]
A pendulum is an object hung from a fixed point that swings back and forth under the action of gravity. In the example of the playground swing, the swing is supported by chains that are attached to fixed points at the top of the swing set. When the swing is raised and released, it will move freely back and forth due to the force of gravity on it. The swing continues moving back and forth without any extra outside help until friction (between the air and the swing and between the chains and the attachment points) slows it down and eventually stops it.
7 0
3 years ago
a projectile is lunched with an initial speed of 60.0mm/s at an angle of 30.0° above the horizontal.The projectile lands on a hi
alexandr402 [8]

Answer:

52 mm/s (approximately)

Explanation:

Given:

Initial speed of the projectile is, u=60.0\ mm/s

Angle of projection is, \theta=30.0\°

Time taken to land on the hill is, t=4\ s

In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

So, the velocity in the horizontal direction always remains the same.

The horizontal component of initial velocity is given as:

u_x=u\cos\theta\\u_x=60\times \cos(30)\\u_x=30\sqrt3\approx52\ mm/s

Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.

6 0
3 years ago
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