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Diano4ka-milaya [45]
3 years ago
9

A firewoman dropped a person onto the safety net. Right before the person hit the net he had a

Physics
1 answer:
sergejj [24]3 years ago
6 0

Answer:

The Mass of a person is calculated to be 28.6kg .

Explanation:

This is based on Kinetic energy and we know, that Kinetic energy is the energy possessed by body by virtue of its motion .

It can be calculated by expression :

K.E=1/2mv²

Velocity of a person = 11.2m/sec

Kinetic energy = 1800J

Mass of person = ?

We know ,

K.E=1/2mv²

so, putting values we have :

1800=1/2 x m x (11.2)²

That is ,

m=1800 x 2 /11.2 x 11.2

or

m=3600/125.44

m = 28.6 kg

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SP1b.
nata0808 [166]

Answer:

2 m/s^2, west

Explanation:

Vf=final velcoity

Vi=initial velocity

t=timw

a =  \frac{vf - vi}{t}

=

\frac{15 - 25}{5}

= - 2 m/s^2

The - changes direction and makes it opposite

2 m/s, west

3 0
3 years ago
As altitude increases in the troposphere and stratosphere, the air temperature does what?
natima [27]

Answer:

1

Explanation:

3 0
3 years ago
The energy emitted from the sun is a product of ________.
Tamiku [17]
Q. The energy emitted from the sun is a product of ________.

A. Fusion
3 0
3 years ago
"Two long parallel wires 24.0 cm apart carry currents of 3.0 A and 8.0 A in the same direction. How far from the wire carrying 3
avanturin [10]

Answer:

6.5454 m

Explanation:

Let the distance from the wire carrying 3 A current is x

Then the distance from the the carrying current 8 A is 24-x

We know that magnetic field due to long wire is given by B=\frac{\mu _0i}{2\pi r}

It is given that magnetic field is zero at some distance so

\frac{\mu _0i_1}{2\pi x}=\frac{\mu _0i_2}{2\pi (24-x)}

Here i_1=3\ A \ and\  i_2=8\ A

So \frac{3}{x}=\frac{8}{24-x}=6.5454\ m

3 0
3 years ago
A diffraction grating contains 15,000 lines/inch. We pass a laser beam through the grating. The wavelength of the laser is 633 n
MatroZZZ [7]

Answer:

Recall the Diffraction grating formula for constructive interference of a light

y = nDλ/w                                      Eqn 1

Where;

w = width of slit = 1/15000in =6.67x10⁻⁵in =   6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m

D = distance to screen  

λ = wavelength of light  

n = order number  = 1

Given  

y1 = ? from 1st order max to the central  

D = 2.66 m  

λ = 633 x 10-9 m  

and n = 1  

y₁ = 0.994m  

Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 1)                =         0.994m

Q b. How far (m) from the central maximum (m = 0) is the second-order maximum (m = 2) observed?

w = width of slit = 1/15000in =6.67x10⁻⁵in =   6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m

D = distance to screen  

λ = wavelength of light  

n = order number  = 1

Given  

y1 = ? from 1st order max to the central  

D = 2.66 m  

λ = 633 x 10⁻⁹ m  

and n =  2

y₂ = 0.994m  

Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 2) =1.99m

8 0
3 years ago
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