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Arte-miy333 [17]
3 years ago
14

What is meant by the statement "Motion is relativel"?

Physics
1 answer:
aleksklad [387]3 years ago
4 0

Answer:

when an object move from its place is called an mation

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A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann
Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/s^{2}

Bring out the given parameters from the question:

Initial Velocity (V_{1}) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

4 0
3 years ago
I need help with this physics question
insens350 [35]
It's impossible to describe WHERE a place is without mentioning ANOTHER place.

... Across the street from -- the bank.
... Next door to -- my house.
... 30 miles west of -- Chicago.
... Up above -- the tree.
... Two days ride out of -- Tulsa.
... Halfway home from -- school.
... Twice as far from Earth as -- the moon is.
... The first seat in -- the second row.
... Behind -- the dog's left ear.
... At the bottom of -- the pool.
... On the tip of -- my tongue.
... In the front seat of -- the car.
... I saw you in -- my dream.
... You're always on -- my mind.

The question is trying to get you to realize that to get from a reference point to a certain position, you have to know

How far
and
In what direction.
4 0
3 years ago
Which space mission visited the moon several times?.
Ivanshal [37]
NASA’s Apollo Mission was the ONLY mission which landed humans on the moon, which it did six times. I’m not very sure about several times, though. Hope this helped!

-chrissydarling
3 0
2 years ago
a bicycle accelerates at 1 m/s² from an initial velocity of 4 m/s² for 10s. Find the distance moved by it during this interval o
BaLLatris [955]

Answer:

90 m

Explanation:

Acceleration, a=\frac {v-u}{t} where v and u are final and initial velocities respectively, t is the time taken

Substituting 1 m/s^{2} for a,  4 m/s for u and 10 s for t then

1*10=v-4

v=14 m/s

From kinematic equations

v^{2}=u^{2}+2as

Making s the subject then

s=\frac {v^{2}-u^{2}}{2a}=\frac {14^{2}-4^{2}}{2\times 1}=90 m

6 0
2 years ago
If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.
Darina [25.2K]

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT  = 15 ^0C = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

  So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

                    c = \frac{2050}{150*10^{-3}*15}  = 911.11 J/kgK

 So object's specific heat = 911.11 J/kgK

5 0
3 years ago
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