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2 years ago

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An astronaut has become detached from her spaceship. She's floating in space holding a wrench. She's 50 m from her current locat

ion. How does she get back to the spaceship?

1 answer:

egoroff_w [7]2 years ago

3 0

Answer:

She throws the wrench in the opposite direction of her destination and the resulting backward force should be enough to get her back to the station.

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A 5.20-N force is applied to a 1.05-kg object to accelerate it rightwards. The object encounters 3.29-N of friction. Determine t

tatiyna

Answer:

Explanation:

We will use the equation F - f = ma, which is just a fancy way of stating Newton's 2nd Law. For us:

F = 5.20 to the right (+)

f = 3.29 to the left (-)

m = 1.05 kg. Therefore,

5.20 - 3.29 = 1.05a and

1.91 = 1.05a so

a = 1.82 m/s/s to the right

8 0

2 years ago

Please help! I especially need help with the second question but help with the first one would be most appreciated!

lara31 [8.8K]

Answer:

a) Team A will win.

b) The losing team will accelerate towards the middle line with 0.01 m/ $s^{2}$

Explanation:

Given that Team-A pulls with a force , $F_{1} = 50N$

and Team-B pulls with a force , $F_{1} = 45N$

∵ $F_{1} > F_{2}$

The rope will move in the direction of force $F_{1}$ .

∴ Team-A will win.

b) Considering both the teams as one system of total mass , $m = 246+253 = 499 kg$

Net force on the system , $F = F_{1} - F_{2}$ = 50-45 = 5N

Applying Newtons first law to the system ,

F = ma , where 'a' is the acceleration of the system.

Since , both the teams are connected by the same rope , their acceleration would be the same.

∴ 5 = 499×a

∴ a = 0.01 m/ $s^{2}$

4 0

3 years ago

At what distance from a long, straight wire carrying a current of 8.7 A is the magnetic field due to the wire equal to the stren

deff fn [24]

Answer:

r= 3.2 cm

Explanation:

Given that

I= 8.7 A

B= 5.4 x 10⁻⁵ T

μo=1.25664 x 10⁻⁶

We know that magnetic filed in wire at a distance r given as

$B=\dfrac{\mu_oI}{2\pi r}$

$r=\dfrac{\mu_oI}{2\pi B}$

By putting the values

$r=\dfrac{1.25664\times 10^{-6}\times 8.7}{2\times \pi \times 5.4\times 10^{-5}}\ m$

r=0.032 m

r= 3.2 cm

5 0

3 years ago

When the net force of opposite forces is zero , the forces are

lorasvet [3.4K]

The answer is balanced

4 0

2 years ago

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 17.3 m/s directly upward.

elena-s [515]

Answer:

h=15.27m

Explanation:

Since at maximum height the vertical velocity must be null it's better to use the formula:

$v_f^2=v_i^2+2ad$

We will use this formula for the vertical direction, choosing the upward direction as the positive one, so we have:

$0=v_i^2+2ah$

or

$h=-\frac{v_i^2}{2a}$

which for our values is:

$h=-\frac{(17.3m/s)^2}{2(-9.8m/s^2)}=15.27m$

7 0

2 years ago