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Archy [21]
2 years ago
14

Pls answerrr it's urgent!!!!!

Physics
1 answer:
lilavasa [31]2 years ago
6 0

Explanation:

the average velocity of the car is 15 m/s example I have this on a test

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encourage all attempts at reading, writing, and speaking

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X = 20 <br> v = 3.5 m/s <br> t = ?
N76 [4]

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16.5 I think

Explanation:

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3 years ago
In a transverse wave the particles Name
7nadin3 [17]

Answer:

The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.

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2 years ago
A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?
vesna_86 [32]

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

v^2=u^2+2as

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

v=u+at

where symbols have the usual meaning

Applying the given values we get

t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds

4 0
3 years ago
An unknown material, m1 = 0.49 kg, at a temperature of T1 = 92 degrees C is added to a Dewer (an insulated container) which cont
erastova [34]

Answer:

c_u=1540.5J/kg^{\circ}K

Explanation:

We know that heat relates to mass, specific heat and variation of temperature experimented because of this heat through the equation Q=mc\Delta T=mc(T_f-T_i). The heat released by the unknown material is absorbed by water, so we have Q_u=-Q_w, and we can write:

m_uc_u(T_{uf}-T_{ui})=-m_wc_w(T_{wf}-T_{wi})

Since thermal equilibrium is reached we know that T_{cf}=T_{wf}=T_f=31^{\circ}C=304^{\circ}K, where we have added 273^{\circ} to convert the temperature from Celsius to Kelvin, as <em>we must do</em>. Since we want the specific heat of the unknown material, we do:

c_u=-\frac{m_wc_w(T_f-T_{wi})}{m_u(T_f-T_{ui})}

Which for our values is:

c_u=-\frac{(1.1kg)(4186J/kg^{\circ}K)((304^{\circ}K)-(294^{\circ}K))}{(0.49kg)((304^{\circ}K)-(365^{\circ}K))}=1540.5J/kg^{\circ}K

3 0
3 years ago
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