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jonny [76]
3 years ago
7

Decide which of the following sports can be the most beneficial in achieving cardiovascular

Physics
1 answer:
Anvisha [2.4K]3 years ago
3 0

If you mean beneficial in improving cardiovascular endurance then the best sports would be swimming, running, basketball, rowing and squash.

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An ice skater spins at 2.5 rev/s when his arms are extended. He draws his arms in and spins at 10.0 rev/s. By what factor does h
Rainbow [258]

Answer:

The moment of inertia decreased by a factor of 4

Explanation:

Given;

initial angular velocity of the ice skater, ω₁ = 2.5 rev/s

final angular velocity of the  ice skater, ω₂ = 10.0 rev/s

During this process we assume that angular momentum is conserved;

I₁ω₁ = I₂ω₂

Where;

I₁ is the initial moment of inertia

I₂ is the final moment of inertia

I_2 = \frac{I_1 \omega_1}{\omega_2} = \frac{I_1*2.5}{10} \\\\I_2 = 0.25I_1 = \frac{1}{4}I_1

Therefore, the moment of inertia decreased by a factor of 4

4 0
3 years ago
What is the speed of a wave with 3Hz frequency and a wavelength of 9 m
SIZIF [17.4K]

Answer:

Given

Frequency (f) = 3Hz

Wavelength = 9 m

Speed = ?

Explanation:

we know

Speed = wavelength * frequency

= 9*3

= 27 m/ s

7 0
3 years ago
A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.
fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force N= mg cos 22

\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N





8 0
3 years ago
Read 2 more answers
A skater of mass 60 kg has an initial velocity of 12 m/s. He slides on ice where the frictional force is 36 N. How far will the
Alexus [3.1K]

Answer:

d = 120 [m]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. Where the energy in the final state (when the skater stops) is equal to the sum of the mechanical energy in the initial state plus the work done on the skater in the initial state.

The mechanical energy is equal to the sum of the potential energy plus the kinetic energy.  As the track is horizontal there is no unevenness, in this way, there is no potential energy.

E₁ + W₁₋₂ = E₂

where:

E₁  = mechanical energy in the initial state [J] (units of Joules)

W₁₋₂ = work done between the states 1 and 2 [J]

E₂  = mechanical energy in the final state = 0

E₁ = Ek = kinetic energy [J]

E₁ = 0.5*m*v²

where:

m = mass = 60 [kg]

v = initial velocity = 12 [m/s]

Now, the work done is given by the product of the friction force by the distance. In this case, the work is negative because the friction force is acting in opposite direction to the movement of the skater.

W₁₋₂ = -f*d

where:

f = friction force = 36 [N]

d = distance [m]

Now we have:

0.5*m*v² - (f*d) = 0

0.5*60*(12)² - (36*d) = 0

4320 = 36*d

d = 120 [m]

7 0
2 years ago
A dog leaps horizontally off a 70 m cliff with a speed of 6 m/s, how far from the base will the dog land?
iris [78.8K]

Answer:

\Delta x=22.67786838m

Explanation:

Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:

y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2 (1)

Where:

y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis

y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis

v_o=initial\hspace{3}velocity

t=travel\hspace{3}time

g=gravity\hspace{3}constant

\theta=Initial\hspace{3}launch\hspace{3}angle

In this case:

\theta=0

Because the dog jumps horizontally

Let's asume the gravity constant as:

g=9.8

y=0

Because when the dog reach the base the height is 0

y_o=70

v_o=6

Now let's replace the data in (1)

y_o-\frac{1}{2} *(9.8)*t^2+70

Isolating t:

t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473

Finally let's find the horizontal displacement using this equation:

\Delta x=v_o*cos(\theta)*t

Replacing the data:

\Delta x=6*1*3.77964473=22.67786838m

8 0
3 years ago
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