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bekas [8.4K]
3 years ago
7

HELP URGENT WILL GIVE BRAINLIEST

Physics
1 answer:
KatRina [158]3 years ago
4 0

Answer:

F • t= m•∆v

Explanation:

the impulse experienced by the object equals the change in momentum of the object. in equation form: F • t= m•∆v

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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

A_{area}=0.140m^{2}\\  p_{air}=1.21 kg/m^{3}\\  m_{mass}=80kg

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

mg=1/2*C*p_{air}*v^{2}*A_{area}\\   v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

v=(115*3.6)km/h\\v=414km/h

5 0
3 years ago
The second part of estate planning consists of:
Akimi4 [234]
Transferring your estate in the manner you have specified 
4 0
3 years ago
Read 2 more answers
What is the definition of physical change
jeka94
They are used to separate mixtures into their component compounds but can no usually be used to deprecate compounds into chemical elements or simpler compounds
7 0
3 years ago
An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric
8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0
3 years ago
As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.01 m l
Valentin [98]

Answer:

The strength of the magnetic field is 75.6\ \mu T.

Explanation:

Given that,

Length of the rod, L = 1.01 m

Speed with which the rod is moving, v = 3.47 m/s

We need to find the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.265 mV across the rod. When the rod is moving with some speed, an emf gets induced and it is given by :

\epsilon=Blv

B is magnetic field

B=\dfrac{\epsilon}{lv}\\\\B=\dfrac{0.265\times 10^{-3}}{1.01\times 3.47}\\\\B=75.6\ \mu T

So, the strength of the magnetic field is 75.6\ \mu T.

6 0
3 years ago
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