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Kobotan [32]
2 years ago
7

An ice hockey puck is tied by a string to a stake in the ice. the puck is then swung in a circle. what force is producing the ce

ntripetal acceleration of the puck?
Physics
2 answers:
Taya2010 [7]2 years ago
8 0
In a circular motion scenario, the force that pulls the revolving object towards the centre is the force that produces the centripetal acceleration. So, in this case, the tension on the string is the force that pulls the puck towards the centre.

Therefore, it is the tension in the string that causes the centripetal acceleration of the puck

Hope I helped!! xx
shtirl [24]2 years ago
3 0

Answer:

In a circular motion scenario, the force that pulls the revolving object towards the centre is the force that produces the centripetal acceleration. So, in this case, the tension on the string is the force that pulls the puck towards the centre.

sampls

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A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i
OLga [1]

Answer:

α = (ω²)/8π

Explanation:

The angular acceleration(α) of the carousel can be determined by using rotational kinematics:

ω² =ωo² + 2αθ

Let's make α the subject of this equation ;

ω² - ωo² = 2αθ

α = (ω² −ωo²)/2θ

Now, from the question, since initially at rest, thus, ωo = 0

Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π

Plugging in the relevant values to get ;

α = (ω²)/2(4π)

α = (ω²)/8π

7 0
2 years ago
An object is thrown horizontally off a cliff with an initial velocity of 5.0 meters per second. the object strikes the ground 3
shutvik [7]
Speed is not a vector so horizontal speed does not sighify anything. If u meant velocity it vill be same as the initial velocity in horizontal direction
7 0
2 years ago
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A massive tractor rolls down a country road. in a perfectly inelastic collision, a small sports car runs into the machine from b
Helga [31]
The answer for this change in the magnitude of momentum is the same for both because momentum is always conserved so both vehicles have the identical change. 
So for determining who has the greater change in kinetic energy, momentum (P) = mv so P^2 = m^2 v^2 P^2 / 2m = 1/2 m v^2 = energy So the weightier the mass the smaller the energy change for the same momentum change so in here, the car has a greater change in kinetic energy.
5 0
2 years ago
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A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s
BlackZzzverrR [31]

Answer:

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

Sf^2=So^2+(2*a*x)

Sf=So+(a*t)

We are going to put our data:

(15m/s)^2=So^2+(2*a*60m)

15m/s=So+(a*6s)

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

\sqrt{(15m/s)^2-(2*a*60m)}=So

15m/s-(a*6s)=So

\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)

[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}

(15m/s)^2-(2*a*60m)}=(15m/s)^{2}-2*(a*6s)*(15m/s)+(a*6s)^{2}

-120m*a=-180m*a+36s^{2}*a^{2}

0=120m*a-180m*a+36s^{2}*a^{2}

0=-60m*a+36s^{2}*a^{2}

0=(-60m+36s^{2}*a)*a

0=a1

\frac{60m}{36s^{2}} = a2

1.67m/s^{2}=a2

If we analyze the situation, we need to have an aceleretarion  greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

Sf=So+(a*t)

15m/s=So+1.67m/s^2*6s

15m/s-(1.67m/s^2*6s)=So

4.98m/s=So

For part c:

We are going to use:

Sf^2=So^2+(2*a*x)

(4.98m/s)^2=0^2+(2*(1.67m/s^2)*x)

\frac{24.80m^2/s^2}{3.34m/s^2}=x

7.42m=x

5 0
2 years ago
A quarterback throws a football at 40km/hr to a receiver 50yd away. How much time does it take the ball to reach the receiver
Akimi4 [234]

Given:

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

To find:

Time required by ball to reach the receiver = ?

Formula used:

speed = \frac{distance}{time}

Solution:

The speed of the ball is given by,

speed = \frac{distance}{time}

Thus,

Time = \frac{distance}{speed}

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

Time = 4.12 second

Hence, ball reaches the receiver in 4.12 second.

3 0
2 years ago
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