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Andrej [43]
3 years ago
8

WILL GIVE BRAINLIEST! QUICK PLEASE HELP!!!

Physics
1 answer:
Fantom [35]3 years ago
4 0

Answer:

mass and location

Explanation:

edgenuity test 100%

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We have a toy gun with a spring constant of 50 N/m. The spring is compressed by 0.2 m. If you neglect friction and the mass of t
Arisa [49]

Answer:

31.6\:\mathrm{m/s}

Explanation:

The elastic potential energy of a spring is given by Us=\frac{1}{2}kx^2, where k is the spring constant of the spring and x is displacement from point of equilibrium.

When released, this potential energy will be converted into kinetic energy. Kinetic energy is given by KE=\frac{1}{2}mv^2, where m is the mass of the object and v is the object's velocity.

Thus, we have:

Us=KE,\\\frac{1}{2}kx^2=\frac{1}{2}mv^2

Substituting given values, we get:

\frac{1}{2}\cdot 50\cdot 0.2^2=\frac{1}{2}\cdot 0.002\cdot v^2,\\v^2=\frac{50\cdot 0.2^2}{0.002},\\v^2=1000,\\v\approx \boxed{31.6\:\mathrm{m/s}}

4 0
3 years ago
How does the elbow medical and lateral epicondylitis?
dlinn [17]
<span>Lateral epicondylitis, or “tennis elbow,” is an inflammation of the tendons that join the forearm muscles on the outside of the elbow. </span>The bony bump on the outside (lateral<span> side) of the </span>elbow<span> is called the </span>lateral epicondyle<span>. The ECRB muscle and tendon is usually involved in </span>tennis elbow<span>. </span><span>
Medial epicondylitis, or “golfer’s elbow,” is an inflammation of the tendons that attach your forearm muscles to the inside of the bone at your elbow. </span>It's identified by pain from the elbow to the wrist on the inside (medial<span> side) of the elbow. The pain is caused by damage to the tendons that bend the wrist toward the palm.</span>
4 0
3 years ago
A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl
Sergeu [11.5K]

Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon m_1 and the mass of the basket m_2, then according to newton's second law:

(1). \:F = (m_1+m_2)a,

where a =0.265m/s^2 is the upward acceleration, and F = 688N is the net propelling force (counts the gravitational force).

Also, the tension T in the rope is 79.8 N more than the basket's weight; therefore,

(2). \:T = m_2g+79.8

and this tension must equal

T -m_2g =m_2a

(3). \:T = m_2g +m_2a

Combining equations (2) and (3) we get:

m_2a = 79.8

since a =0.265m/s^2, we have

\boxed{m_2 = 301.13kg}

Putting this into equation (1) and substituting the numerical values of F and a, we get:

688N = (m_1+301.13kg)(0.265m/s^2)

\boxed{m_1 = 2295 kg}

Thus, the mass of the balloon and the basket is  2295 kg and 301 kg respectively.

8 0
3 years ago
The _______ is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument. A.
Vladimir [108]

"A is correct answer." The effective length of the tube is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument. "Hope this helps!" "Have a great day!" "Thank you for posting your question!"

8 0
3 years ago
Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
3 years ago
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