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marusya05 [52]
3 years ago
10

A player kicks a football at an angle of 30.0 above the horizontal. The football has an initial velocity of 20.0 m/s. Find the h

orizontal component of the velocity and the maximum height attained by the football.
a. 10.0 m/s, 17.6 m
b. 17.3 m/s, 5.10 m
c. 25.1 m/s, 7.40 m
d. 30.3 m/s, 9.50 m ido
Physics
1 answer:
Verizon [17]3 years ago
6 0

Answer:

Option b) 17.3 m/s and 5.10 m

Explanation:

For this case, we can consider a parabolic shot, for which we have an initial speed of 20 m/s, which has a vertical component (Voy) and a horizontal component (Vox)  - see attached picture 1 on this case -

According to the figure, we have a right rectangle, for which we can obtain the following formulas:

V₀ₓ= V₀Cosα

V₀y=V₀Senα

With these formulas, we can first obtain the horizontal component of velocity:

V₀x=20xCos30= 17.32 m/s

Then we need to obtain the maximum height (point C on the chart). At this point, we can think about a freefall, where speed is given by :

V=gt , where g= gravity acceleration  and t=time

Our maximun time at point C will be: tmax=V/t, and we can replace vertical component of speed, given by V₀Senα, then:

Tmax=V₀Senα/g

Finally, if we still consider a movement in free fall, time is given by :

t= √(2h/g)   then h=(gt²)/2 and replacing our time previously found:

hmax=(V₀xsenα)²/2g

And replacing our numbers: hmax= 5.10 m

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A car accelerates for 10 seconds. During this time, the angular
bagirrra123 [75]

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, \omega_i = 10 rad/s

final angular velocity, \omega_f = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2

Therefore, the angular acceleration of the car is 1.5 rad/s²

4 0
3 years ago
Which vector shows the velocity of the yo-yo at this moment?
Helga [31]

The vector which shows the velocity of the yo-yo at this moment is A and is denoted as option A.

<h3>What is Velocity?</h3>

This is referred to the rate of change of distance with time and the unit is metre per seconds.

A shows the direction of motion is same as the direction of velocity and is the tangent at that particular point hence why it was chosen as the most appropriate choice.

Read more about Velocity here brainly.com/question/6504879

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6 0
2 years ago
Now it's your turn
eduard

Answer:

The average force the golf club exerts on the ball is 600 N

Explanation:

Newton's second law of motion states that force, F, is directly proportional to the rate of change of momentum produced

F = m× (v₂ - v₁)/(Δt)

The given parameters of the motion of the ball are;

The mass of the ball, m = 45 g = 0.045 kg

The initial velocity of the ball, v₁ = 0 m/s

The speed with which the ball was hit by the golfer, v₂ = 40 m/s

The duration of contact between the golf club and the ball, Δt = 3 ms = 0.003 seconds (s)

By Newton's law of motion, the average force, 'F', which the golf club exerts on the ball is therefore, given as follows;

F = 0.045 kg × (40 m/s - 0 m/s)/(0.003 s) = 600 N

The average force the golf club exerts on the ball = F = 600 N.

5 0
3 years ago
Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that
Greeley [361]

Answer:

The  value  is   m_i =  0.0234 \  kg

Explanation:

Generally from the calorimetry principle we have that

      Heat \ loss\ by\ ice  =  heat\ gained\ by\  water\

So here heat gained water is mathematically represented as i.e

      Q_w  = m_w  *  c_w *  (T_w - T )

substituting  0.150 kg for m_w , 4200 J/kg.°C for  c_w , 100°C for   T_w and  75°C for  T

We have  

       Q_w  = 0.150  *  4200 *  (100 - 75 )

         Q_w  =15750 \  J

The Heat loss by the ice is mathematically represented as

      Q_i  = Q_1 + Q_2 +  Q_3

Here     Q_1 is the energy to move the ice to its melting point which is evaluated as  

        Q_1  =  m_i *  c_i * ( T_o -T_i)

Here  m_i is the mass of  ice

       c_i is the specific heat of ice with value  2.05 * 10^3   J/kg.°C

          T_o temperature of ice at melting point with value 0°C

           T_i is the temperature of ice with value  -12°C

Q_2 is the energy to move the ice from its  its melting point to liquid which is evaluated as  

     Q_2  = m_i  *  L

        Here  L  is the Latent heat of melting of ice with value    334 * 10^3   J/kg

Q_3 is the energy to move the ice from  liquid  to  the equilibrium temperature  which is evaluated as        

       Q_1  =  m_i *  c_w * ( T -T_o)

So  

     Q_i  = m_i [ c_i * ( T_o -T_i) + L  + c_w * ( T -T_o) ]

=>   Q_i  = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3  +  4200 * ( 75 - 0) ]

From

 Heat \ loss\ by\ ice  =  heat\ gained\ by\  water\

We have that

  m_i *  673600  =15750

=>     m_i =  \frac{15750}{673600}

=>     m_i =  0.0234 \  kg

8 0
4 years ago
What is the maximum mass that can hang without sinking from a 20-cm diameter Styrofoam sphere in water? Assume the volume of the
alex41 [277]

Answer:

the maximum mass that can hang without sinking is 2.93 kg

Explanation:

Given: details:

sphere diameter  d = 20 cm

so, radius r = 10 cm  = 0.10 m

density of the Styrofoam sphere D = 300 kg/m3

sphere volume  V = \frac{4}{3} \pi r^3

                                                   =\frac{4}{3} \pi 0.10^3

                                                   =4.18*10^{-3} m^3

we know that

Density = \frac{Mass}{Volume}

mass  M = Density * Volume

                                  = (300)(4.18*10^{-3} m3)

                                  =1.25 kg

mass of the water displace = volume *density  of water

                                                 = 4.18*10^{-3} m3 * 1000

                                                 = 4.18 kg

The difference between the mass of water and mass of styrofoam is the amount of mass that the sphere can support

=4.18 kg  -1.25 kg

= 2.93 kg

3 0
4 years ago
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