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o-na [289]
3 years ago
5

A ball is thrown upward.

Physics
1 answer:
valentina_108 [34]3 years ago
8 0

Answer:

It is 10.75

Explanation:

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What actions can be explained by physics?
Vesna [10]
Всяко действие има равно по големина и противоположно по посока противодействие.
7 0
4 years ago
A 91.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.73 r
KIM [24]

Answer:

\omega_f = 1.08 rad/s

Explanation:

As we know that there is no external torque on the system of platform, banana and the monkey

so the angular momentum of the system will remains conserved

so here we can say

L_i = L_f

I_o\omega = (I_o + m_1r_1^2 + m_2r_2^2)\omega_f

here we know that

I_o = \frac{1}{2}mR^2

I_o = \frac{1}{2}(91.1)(1.61)^2 = 118.1 kg m^2

m_1 = 9.41 kg

r_1 = \frac{4}{5}(1.61) = 1.29 m

m_2 = 21.1 kg

Now from above equation

118.1(1.73) = (118.1 + 9.41(1.29^2) + 21.1(1.61^2))\omega_f

118.1(1.73) = 188.45\omega_f

\omega_f = 1.08 rad/s

8 0
3 years ago
I need to know the correct answer please ?
mrs_skeptik [129]

Answer:

675 Pa.

Explanation:

F = 5+2cos(15t) kN

Area (a) = 8*10-3 m2

Now at t =4 sec

            F= 5+2cos(60)

             = 5+2*0.5

              = 6  kN

Now ,force efficiency is 90%.

Hence,the effectively transmitted force,

            Fe = 0.90*6

                = 5.4 kN

Hence,pressure is given as,

            P = Fe/a

                = 5.4*10^3/(8 *10^-3))

              P = 675 Pa....answer

6 0
4 years ago
A cyclist racing in the keiran is riding at the top of the track at 5m/s. Then he sprints downhill in the sprinting to the finis
kumpel [21]
The cyclist is moving by uniformly accelerated motion, with an initial velocity of v_i=5~m/s and an acceleration of a=9~m/s^2. 
The acceleration is given by
a= \frac{v_f-v_i}{\Delta t}
where v_f is the final velocity and \Delta t is the time between the end and the beginning of the motion, and in our case it is 1.75 s. Therefore, from this relationship we can find the final velocity:
v_f=v_i + a \Delta t=5~m/s+9~m/s^2 \cdot 1.75~s=20.75~m/s
6 0
4 years ago
The first charge is pulling on the second. Is the second pulling on the first? Explain your reasoning
Alex73 [517]

Answer:

law of action and reaction.

Explanation:

In Newton's three laws it is established that forces act in pairs, if one body interacts with another the second interacts with the first, this is the so-called law of action and reaction.

In this case, when the first load pulls on the second, the second pulls on the first, the two forces are not canceled because each one is applied to a different body.

Therefore the magnitude of the forces is the same, but the direction is opposite and each one is applied in a body

7 0
3 years ago
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