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Neko [114]
4 years ago
12

What property of matter must be overcome in order for the gummy bear to shoot out of the cup?

Physics
1 answer:
ddd [48]4 years ago
7 0

Answer:

Propulsion.

Explanation:

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2. An athlete of average size is hanging from the end of a 20 m long rope, which has a mass of 4 kg and is attached to a hook in
a_sh-v [17]

Answer:

  t = 0.319 s

Explanation:

With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by

             v = √T/λ

Linear density is

           λ = m / L

           λ = 4/20

           λ = 0.2 kg / m

The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg

           T = W = mg

           T = 80 9.8

           T = 784 N

The pulse rate is

          v = √(784 / 0.2)

          v = 62.6 m / s

The time it takes to reach the hook can be searched with kinematics

          v = x / t

          t = x / v

          t = 20 / 62.6

          t = 0.319 s

7 0
3 years ago
Find the approximate value of the circumference of a circle with the given radius. Use π = 3.14. Round your results to one more
lara [203]
In order to find the circumference of a circle you need to use the equation C=2πr.
3 0
3 years ago
Uses of Photodiode.State the 8 uses of phototide...................​
Inessa05 [86]

Answer:

Photodiodes are used in consumer electronics devices such as compact disc players, smoke detectors, medical devices and the receivers for infrared remote control devices used to control equipment from televisions to air conditioners. For many applications either photodiodes or photoconductors may be used.

6 0
3 years ago
A 310-g air track cart is traveling at 1.25 m/s and a 260-g cart traveling in the opposite direction at 1.33 m/s. What is the sp
UNO [17]

Answer:

v_{CM}=0.0732\ m/s

Explanation:

given,

mass of the cart 1, m₁ = 310 g

speed of car 1 , v₁ = 1.25 m/s

mass of cart 2, m₂ = 260 g

speed of cart 2, v₂ = -1.33 m/s

speed of center of mass

v_{CM}=\dfrac{m_1v_1 + m_2 v_2}{m_1 + m_2}

v_{CM}=\dfrac{0.31\times 1.25 +0.26\times (-1.33)}{0.31+0.26}

v_{CM}=\dfrac{0.0417}{0.57}

v_{CM}=0.0732\ m/s

Hence, speed of center of mass of the system is equal to 0.0732 m/s

7 0
4 years ago
A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the
WITCHER [35]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\

<u>v_f = 10.85 m/s</u>

8 0
3 years ago
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