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love history [14]
2 years ago
15

Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th

seconds​
Physics
1 answer:
krek1111 [17]2 years ago
7 0

Answer:

90

Explanation:

You might be interested in
a block of wood has a length of 4 cm a width of 5 cm and a height of 10 cm what is the volume of the wood
geniusboy [140]

Volume= Length X width X height.

Plug in the values for each and solve for the volume.

V= (L)(W)(H)

V=(4cm)(5cm)(10cm).


8 0
3 years ago
If a block of wood dropped from a tall building has attained a velocity of 78.4 m/s how long has it been falling
ryzh [129]

Gravity adds 9.8 m/s to the speed of a falling object every second.

An object dropped from 'rest' (v = 0) reaches the speed of 78.4 m/s after  falling for (78.4 / 9.8) = <em>8.0 seconds</em> .

<u>Note:</u>

In order to test this, you'd have to drop the object from a really high cell- tower, building, or helicopter.  After falling for 8 seconds and reaching a speed of 78.4 m/s, it has fallen 313.6 meters (1,029 feet) straight down.

The flat roof of the Aon Center . . . the 3rd highest building in Chicago, where I used to work when it was the Amoco Corporation Building . . . is 1,076 feet above the street.

5 0
3 years ago
A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca
igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

f=mg\ sin\theta

f=1100\ kg\times 9.8\ m/s^2\ sin(4)

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0
3 years ago
A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa
Darya [45]

Answer:

x=0.53x10^{-3} m

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1.x10x^{-6} \frac{C}{m^{2}}

ε=8.85x10^{-12} \frac{C^{2}}{n*m^{2}}

E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}

Force of charge is

F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N

F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2}  \\x_{f}=0.53x^{-3}m

5 0
2 years ago
Show all work.
lys-0071 [83]

The new gravitation force at the new location is 40 N

Explanation:

The weight of the astronaut is given by the equation

F=mg (1)

where

m is the mass of the astronaut

g is the acceleration of gravity

The acceleration of gravity at a certain distance r from the centre of the Earth is given by

g=\frac{GM}{r^2}

where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as

F=\frac{GMm}{r^2}

When the astronaut is on the Earth's surface, r=R (where R is the Earth's radius), so his weight is

F=\frac{GMm}{R^2}=640 N

Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

r'=3R+R=4R

Therefore, the new weight is

F'=\frac{GMm}{(4R)^2}=\frac{1}{16}\frac{GMm}{R^2}=\frac{F}{16}

Which means that his weight has decreased by a factor 16: therefore, the new weight is

F'=\frac{640}{16}=40 N

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

3 0
3 years ago
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