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3 years ago

I lost 1 point on this and I don’t know what I did wrong.

Physics

2 answers:

Lilit [14]3 years ago

8 0

-- there's no normal force acting on the diving falcon

-- drag IS kinetic friction; both terms refer to the same single force

-- it might be a bit more elegant if you referred to a force as "gravity" instead of "weight"

Viktor [21]3 years ago

4 0

Answer:

i think its because u gave the almost every answer the same exact thing. all the questions have different ways of moving which means different forces for each one i hope this helps :)

Explanation:

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What is the gravitational force between Mars and the sun? 7.43 × 1030 N 1.79 × 1026 N 1.65 × 1021 N 3.76 × 1032 N

VMariaS [17]

The gravitational force between Mars and the Sun is $1.65\cdot 10^{21} N$

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we have:

$m_1 = 1.99\cdot 10^{30} kg$ is the mass of the Sun

$m_2 = 6.39\cdot 10^{23} kg$ is the mass of Mars

$r=229\cdot 10^6 km = 229\cdot 10^9 m$ is the average distance Mars-Sun

Substituting into the equation, we find the gravitational force:

$F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(6.39\cdot 10^{23})}{(229\cdot 10^9)^2}=1.62\cdot 10^{21} N$

So, the closest answer is

$1.65\cdot 10^{21} N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea

viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

$L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}$ (1)

time taken = 5 years

We know that :

time = $\frac{distance}{speed}$

5 = $\frac{L''}{v}$ (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

$\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

$\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

$t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}$

Solving the above eqn:

t'' = 20.56 years

4 0

3 years ago

Chris walks 25 m in the positive direction on a number line, then turns around and walks 15 m in the opposite direction. What is

7nadin3 [17]

Answer:

$|d| = |(25 - 15)| \\ |d| = 10 \: m$

3 0

3 years ago

When practicing an oral presentation what can you do to prepare for the presentation

Igoryamba

When practicing an oral presentation, you can prepare by writing a draft and practice reading aloud what you are going to say before your oral presentation.

8 0

3 years ago

Find the force between 2C and -1C separated by a distance 1m. Indicate if it is an attractive force (negative) or a repulsive fo

telo118 [61]

Answer:

Explanation:

Its definitely an Attractive force since the two charges are Unlike.

From Coulombs Law

F=kq1q2/R²

Given

K=9x10^9

R=1m

q1=2C

q2=-1C

F=(9x10^9 x 2 x -1)/1²

F= - 1.8x10^10N. (Attractive).

6 0

3 years ago