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Sveta_85 [38]
3 years ago
8

Which has the greater change in momentum, a 50 gram clay ball that strikes a wall at 1 m/s and sticks or a 50 gram superball tha

t strikes the wall at 1 m/s and bounces away at 0.8 m/s. Explain?
Physics
1 answer:
jek_recluse [69]3 years ago
5 0

Answer: 50 gram superball that strikes the wall at 1 m/s and bounces away at 0.8 m/s has greater change in kinetic energy.

Explanation:

50 gram superball that strikes the wall at 1 m/s and bounces away at 0.8 m/s has the greater change in kinetic energy because the collision is elastic in nature that is bodies separates after collision and doesn't lose any kinetic energy.

Also for an elastic collision, both the momentum and energy of the bodies are conserved compare to inelastic collision where only momentum is conserved but not the kinetic energy(this is attributed to bodies that sticks together after collision).

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In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr
spin [16.1K]

Answer:

a) v_p=9.35m/s

Explanation:

From the question we are told that:

Open cart of mass   M_o=50.0 kg

Speed of cart   V=5.00m/s

Mass of package   M_p=15.0kg

Speed of package at end of chute V_c=3.00m/s

Angle of inclination   \angle =37

Distance of chute from bottom of cart   d_x=4.00m

a)

Generally the equation for work energy theory is mathematically given by

  \frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2

Therefore

  \frac{1}{2}u^2+gh=\frac{1}{2}v_p^2

  v_p=\sqrt{2(\frac{1}{2}u^2+gh)}

  v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}

  v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}

  v_p=9.35m/s

4 0
3 years ago
Item 4 Which conditions produce the smallest and largest ocean waves? Choose the two correct answers.
Darya [45]

Answer:

strong winds that blow for a long time over a great distance

weak winds that blow for short periods of time with a short fetch

Explanation:

When the winds are weak and blow for short periods, we experience the smallest ocean waves but when there are strong winds over a longer duration, the largest ocean waves are seen. Therefore, the conditions to produce the smallest and largest ocean waves are strong winds that blow for a long time over a great distance and weak winds that blow for short periods of time with a short fetch.

5 0
3 years ago
Read 2 more answers
If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 1030 kg, what is the average distance between the planet a
Levart [38]

Answer:

9.36*10^11 m

Explanation

Orbital velocity  v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

5 0
3 years ago
male lions and human sprinters can both accelerate at about 10.0 m/s2. if a typical lion weighs 170 kg and a typical sprinter we
Rasek [7]

The difference in the force exerted by the ground during a race between these two species is 259.25

<h3>What is force?</h3>

Forces are influences that can change the movement of an object. A force can change the velocity of an object with mass, i.e. accelerate it. Forces can also be described intuitively by pushing or pulling. A force has both magnitude and direction and is a vector quantity.

Force exerted by the lion:

Normal force = mg

170 = m × 9.8

m = 17.34 kg

Forward force = ma

= 17.34 × 10.0

= 173.4 N

Total force = Normal force + Forward force

Total force = 170 + 173.4

= 343.4 N

Force exerted by the man:

Normal force = mg

75 = m × 9.8

m = 7.65 kg

Forward force = ma

= 7.65 × 10.0

= 76.5 N

Total force = Normal force + Forward force

Total force = 7.65 + 76.5

= 84.15 N

The difference in the forces exerted by these species is:

343.4 N - 76.5 N = 259.25 N

To know more about force visit:

brainly.com/question/19734564

#SPJ4

3 0
1 year ago
When 108 g of water at a temperature of 22.5?
FrozenT [24]
<span>ΔT for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC. Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used q=mCΔT 11466.58 Joules = 65.1g x 4.18 J / g C x ΔT 11466.58/(65.1gx4.18)=ΔT ΔT=42.14oC So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC ΔT of second sample).</span>
3 0
4 years ago
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