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Hunter-Best [27]
3 years ago
13

Help asap physics homework, i'll give brainliest

Physics
2 answers:
Georgia [21]3 years ago
7 0

Answer:

a= 0.2 m/s^2\\s  = 20m\\t =  2 s

Explanation:

gievn

two forces are given

f_{1} = 2000 N \\f_{2} = 1800 N\\F_{net} = 2000 -1800  = 200N

A) (a) What is the acceleration of the 1000-kg boat?

f_{net} = ma\\a  = \frac{f_{net}}{m}

     a = \frac{200}{1000}\\a= 0.2 m/s^2

B) If it starts from rest, how far will it move in 10.0 s?

using kinematic equation

s = ut +\frac{1}{2}at^2\\ u = 0  \   m/s\\\\S = \frac{1}{2} at^2 \\s = \frac{1}{2} \times 0.2 \times (10)^2\\s =  10 m

(c) What will its velocity be at the end of this time?

v= u +at \\u = 0 m/s \\v =  0.2\times 10\\v = 2 m/s

Bess [88]3 years ago
5 0

Answer:bippity boppity yee

Explanation:

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Imagine that the apparent weight of the crown in water is Wapparent=4.50N, and the actual weight is Wactual=5.00N. Is the crown
9966 [12]

Answer:

Explanation:

Actual weight, Wo = 5 N

Apparent weight, W = 4.5 N

density of water = 1 g/cm^3 = 1000 kg/m^3

density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3

Buoyant force = Actual weight - Apparent weight

Volume x density of water x g = 5 - 4.5

V x 1000 x 9.8 = 0.5

V = 5.1 x 10^-6 m^3

Weight of gold = Volume of gold x density of gold x gravity

W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N

As W' is less than W so, it is not pure gold.

4 0
3 years ago
What is the source of the radioactive nuclei present in spent fuel rods?
Debora [2.8K]
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3 years ago
What is the slowest type of radiation?
tekilochka [14]
Velocity is the answer
3 0
3 years ago
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
By including
Elena-2011 [213]

Answer:

Option D.

Data; changing attitudes, beliefs, or values

Explanation:

When a speaker adds statistical data to prove the point, the audience's perception about things changes since the data implies that research has been conducted hence reliable. Consequently, the attitude of people and beliefs may change when such data is presented to them. Therefore, option D is correct.

3 0
3 years ago
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