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3 years ago

Ram jumps onto a cement floor from a height of 1m and comes to rest in 0.1sec.

Physics

1 answer:

umka2103 [35]3 years ago

5 0

Answer:

3/10 F.

Explanation:

Height ( h ) = 1m

Time taken ( t ) = 0.1 second

Height² ( h² ) = 9m

Time taken² ( t² ) = 1 second

Solution,

F = ma

= m ( v - u ) / t

= m √2gh / t

now,

F/F² = √h/h² × t/t²

F¹ = 3/10 F.

answer !!

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18 kilogram Mass Blokus addressed a level surface if the coefficient of static friction between the Block in the surface is 0.6

Tasya [4]

Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what horizontal force is required to just move the blcok ( take gravity as 10m/s2 )

Answer:

108 N

Explanation:

From the question,

Applying

F' = mgμ................ Equation 1

Where F' = Frictional force = horizontal force required to just move the block, m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.

From the question,

Given: m = 18 kg, μ = 0.6, g = 10 m/s²

Substitute these values into equation 1

F' = 18×0.6×10

F' = 108 N

4 0

3 years ago

Consider two antennas separated by 9.00 m that radiate in phase at 120 MHz, as described in Exercise 35.3. A receiver placed 150

alexgriva [62]

Answer:

$\phi=4.52 rad$

Explanation:

From the question we are told that

Distance b/e antenna's $d=9.00m$

Frequency of antenna Radiation $F_r=120 MHz \approx 120*10^6Hz$

Distance from receiver $d_r=150m$

Intensity of Receiver $i= 10$

Distance difference of the receiver b/w antenna's $(r^2-r^1)=1.8m$

Generally the equation for Phase difference $\phi$ is mathematically given by

$\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)$

$\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8$

$\phi=\frac{4\pi}{5} *1.8$

Therefore phase difference f between the two radio waves produced by this path difference is given as

$\phi=4.52 rad$

7 0

3 years ago

A proton moves at constant velocity in the direction, through a region in which there is an electric field and a magnetic field.

dimaraw [331]

Answer:

$F_{net}=0$

Explanation:

It is given that, a proton moves at constant velocity, through a region in which there is an electric field and a magnetic field such that,

The electric field is, E = 800 V/m

Magnetic field, B = 0.25 T

We know that the net force in the region of magnetic and electric field is given by Lorentz forces. But here, the proton moves with constant velocity. So, the net force acting on it is 0.

i.e.

$F_{net}=0$

Hence, this is the required solution.

7 0

3 years ago

Physics formula for Celsius to Fahrenheit

worty [1.4K]

F=9/5C+32 is the formula for Celsius to Fahrenheit.

7 0

4 years ago

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