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lara31 [8.8K]
3 years ago
14

if I spacecraft seems to be Motionless in deep space is given some type of quick push what will happen

Physics
1 answer:
ss7ja [257]3 years ago
7 0
When we give a quick push to the spacecraft, we apply a force in a short time. This is equivalent to saying that we give an impulse to the spacecraft. As the momentum theorem says the impulse that we give to the spacecraft is equal to the change in its momentum
(F/\Delta t) = \Delta P
where the momentum is defined as the product between the mass and velocity.
P =mv
Therefore when we apply a quick push the speed of the spacecraft will change from zero to a constant nonzero value.

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According to a college survey, 22% of all students work full time. find the mean for the number of 3) students who work full tim
attashe74 [19]

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E[x] = np = (16)(0.22) = 3.52 
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3 0
3 years ago
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

Atmospheric pressure P_{atm}= 1.013\times10^{5}\ Pa    

drop height h'= 27.1 mm

Density of mercury \rho= 13.59 g/cm^3

We need to calculate the height

Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

h=0.76\ m

We need to calculate the new height

h''=h - h'

h''=0.76-27.1\times10^{-3}

h''=0.76-0.027

h''=0.733\ m

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Using formula of atmospheric pressure

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Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

P=0.97622\times10^{5}\ Pa

Hence, The atmospheric pressure is 0.97622\times10^{5}\ Pa.

7 0
3 years ago
Please, Help!!
Elenna [48]

Let, 1st force = a

2nd force = b

A.T.Q,

a+b = 10

a-b = 6

Calculate for a & b, you'll get a=8 & b= 2

After increasing by 3, it'll be a = 8+3 = 11 & b=2+3 = 5

Resultant force at 90 degree angle = 11+5 = 16 Newtons

7 0
3 years ago
Read 2 more answers
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Ksju [112]
Because of the hint we can conclude what equation we need to solve this problem. We have power and duration that means that we need to express energy:

1 joule = 1watt * 1 second
or
E (energy) = P (power) * t (time duration)
E = 350 * 30 = 10500 joules.
7 0
3 years ago
A moving roller coaster speeds up with constant acceleration for 2.3\,\text{s}2.3s2, point, 3, start text, s, end text until it
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Answer:

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Explanation:

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0

​  

v, start subscript, 0, end subscript of the roller coaster.

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