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3 years ago

a body is projected vertically upward with velocity u returns to starting points in 4 sec find the value of u if g=10m/s

Physics

1 answer:

Lesechka [4]3 years ago

7 0

Answer:

u = 39.2 m/s

Explanation:

use v = u — gt. since g is negative upwards.

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A sound signal traveling underwater has a frequency of 230 hertz. The speed of sound underwater is 1.45 × 103 meters/second. Wha

Sholpan [36]

Wavelength = (wave speed) / (frequency)

= (1,450 m/s) / (230 / sec)

= 6.30 meters (rounded)

Hope this helped friend! :)

4 0

3 years ago

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In a photoelectric effect experiment you illuminate a metal with light of an unknown wavelength and measure the maximum kinetic

grigory [225]

Answer:

(a). The first wavelength is 401.0 nm.

(b). The metal's work function is 2.55 eV.

Explanation:

Given that,

Maximum kinetic energy = 0.65 eV

Second wavelength $\lambda_{2}= \dfrac{2}{3}\times\lambda_{1}$

(a). We need to calculate the wavelength

Using equation of work function for first wavelength

$\dfrac{hc}{\lambda_{1}}-W_{0}=0.65\ eV$ .....(I)

For second wavelength,

$\dfrac{hc}{\lambda_{2}}-W_{0}=2.3\ eV$

Put the value of second wavelength

$\dfrac{1.5 hc}{\lambda_{1}}-W_{0}=2.3\ eV$ ....(II)

By subtraction equation (I) from (II)

$0.5\dfrac{hc}{\lambda_{1}}=1.55$

$\lambda_{1}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3.1\times1.6\times10^{-19}}$

$\lambda_{1}=4.010\times10^{-7}\ m$

$\lambda_{1}=401.0\times10^{-9}\ m$

$\lambda_{1}=401.0\ nm$

(b). We need to calculate the work function

Using formula of work function

$W_{0}=(\dfrac{hc}{\lambda}-0.55)$

Put the value into the formula

$W_{0}=(\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.6\times10^{-19}\times401.0\times10^{-9}}-0.55)$

$W_{0}=2.55\ eV$

Hence, (a). The first wavelength is 401.0 nm.

(b). The metal's work function is 2.55 eV.

4 0

3 years ago

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Two blocks are sliding along a frictionless track. Block A (mass 4.03 kg) is moving to the right at 3.00 m/s. Block B (mass 4.84

julsineya [31]

Answer:

The total momentum of the system before the collision is 5.334 kg-m/s towards left.

Explanation:

Given that,

Mass of the block A, $m_A=4.03\ kg$

Speed of block A, $v_A=3\ m/s$

Mass of the block B, $m_B=4.8\ kg$

Mass of block B, $u_B=-3.6\ m/s$

Let p is the total momentum of the system before the collision. It is given by :

$p=m_Av_A+m_Bv_B\\\\p=4.03\times 3+4.84\times (-3.6)\\\\p=-5.334\ kg-m/s$

So, the total momentum of the system before the collision is 5.334 kg-m/s towards left. Hence, this is the required solution.

6 0

3 years ago

PLEASE HELP ASAP!!! PLEASEEEEEEE

Andrew [12]

-- Top picture:

'a' and 'b' have the same magnitude, and both are positively charged.

-- Bottom picture:

The charge on the particle is negative.

8 0

3 years ago

Which object would be the best insulator of electricity?

solniwko [45]

Answer:

a rubber band

Explanation:

hope this helps!

7 0

3 years ago

Read 2 more answers