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3 years ago

Destinations humans have traveled in space

Physics

1 answer:

raketka [301]3 years ago

8 0

Yes humans have traveled in space before

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A dense metal sphere is dropped from a 10-meter tower, and at the exact same time an identical metal sphere is thrown horizontal

givi [52]

Answer:

A. Both spheres land at the same time.

Explanation:

The horizontal motion doesn't affect the vertical motion. Since the two spheres have the same initial vertical velocity and same initial height, they land at the same time.

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3 years ago

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A student pushes against a wall with a force of 30N. The wall does not move. What amount of force does the wall exert on the stu

True [87]

Answer:

Explanation:

they both have to be the same for both to not move

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2 years ago

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What is the final velocity of a car that starts at 22 m/s and accelerates at 3.78 m/s for distance of 45 m

Pepsi [2]

v^2 = v0^2 +2ad v^2 = 22^2 + 2*3.78*45 = 824.2 v= √824.2 = 28.7 m/s

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3 years ago

A beryllium-9 ion has a positive charge that is double the charge of a proton, and a mass of 1.50 ✕ 10−26 kg. At a particular in

seropon [69]

Answer:

Magnetic force, $F = 3.52\times 10^{-13}\ N$

Explanation:

Given that,

A beryllium-9 ion has a positive charge that is double the charge of a proton, $q=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C$

Speed of the ion in the magnetic field, $v=5\times 10^6\ m/s$

Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.

The magnitude of the field is 0.220 T.

We need to find the magnitude of the magnetic force on the ion. It is given by :

$F=qvB\\\\F=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\\\\F=3.52\times 10^{-13}\ N$

So, the magnitude of magnetic force on the ion is $3.52\times 10^{-13}\ N$ .

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3 years ago

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A Person whose weight is 5.20 x 10^2 N is being pulledup

Gnoma [55]

Answer:

Explanation:

Given

Weight of Person $W=5.20\times 10^{2} N$

Cave is $h=35.1 m$ deep

Breaking stress $T=569 N$

Net Force on Person

$F_{net}=569-520=49 N$

$a_{net}=\frac{F_{net}}{\frac{W}{g}}$

$a_{net}=\frac{49}{\frac{520}{9.8}}$

$a_{net}=0.923 m/s^2$

The shortest time such that the person can be taken out of cave

$h=ut+\frac{1}{2}at^2$

where

h=distance moved

t=time

a=acceleration

$35.1=0+\frac{1}{2}(0.923)(t)^2$

$t^2=76.05$

$t=\sqrt{76.05}$

$t=8.72\ s$

6 0

3 years ago