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ivanzaharov [21]
3 years ago
11

Look at the graph and extrapolate how many baskets were made by the team at 8.5 minutes.

Physics
1 answer:
scoundrel [369]3 years ago
8 0
C
djjsksksndndjjsbdndnnamqkandbdbs
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a 3 kg piece of putty that is moving with a velocity of 10 m/s collides and sticks to an 8 kg bowling ball that was at rest. wha
defon

The final velocity is 2.7 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.

Therefore we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 3 kg is the mass of the putty

u_1 = 10 m/s is the initial velocity of the putty (we take its direction as positive direction)

m_2 = 8 kg is the mass of the ball

u_2 = 0 m/s is the initial velocity of the ball (at rest)

v is the final combined velocity of the two putty+ball

Re-arranging the equation and substituting the values, we find the  final combined velocity:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(3)(10)+0}{3+8}=2.7 m/s

And the positive sign indicates their final direction is the same as the initial direction of the putty.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

3 0
3 years ago
Something that flows easily, like water
Yanka [14]
Sweat and blood flow easily too
6 0
4 years ago
Velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on applying a force continuously for 2 s. Find out the force ap
Alenkasestr [34]

Answer:

Force=10N

Velocity=13m/s

Explanation:

Given,

Mass = 5kg

initial velocity=3m/s

Final velocity=7m/s

Time=2s

Now acceleration= v-u/t

=7-3/2

=4/2

=2m/s²

So, acceleration=2m/s²

Now,

F=ma

=5*2

=10N

7 0
3 years ago
In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact
OLga [1]

Answer:

v=3.73m/s,   a=3.35m/s^{2}

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

K_{A}+U_{Ag}=U_{Cs}+W_{Cf}

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}

so we can now solve this for k, which is the constant of the spring.

k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}

and now we can substitute the respective data:

k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}

and we can solve for the Velocity in B, so we get:

V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}

we can now input all the data directly, so we get:

V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}

which yields:

V_{B}=3.73m/s

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

\sum F=ma

so we get:

F_{s}+f-W=ma

Supposing the acceleration is positive when it points upwards.

So we can solve that for the acceleration, so we get:

a=\frac{F_{s}+f-W}{m}

or

a=\frac{ky_{B}+f-W}{m}

and now we substitute the data we know:

a=\frac{(9300N/m)(1m)+(17000N)-(2000kg)(9.8m/s^{2})}{2000kg}

which yields:

3.35m/s^{2}

8 0
4 years ago
What two things do you need to know to describe the velocity of an object?
pickupchik [31]
The distance it traveled and the time that it took to travel that distance
3 0
3 years ago
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