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Mkey [24]
3 years ago
13

GIVING BRAINLIEST PLEASE HELP!!

Physics
2 answers:
Leviafan [203]3 years ago
6 0
A and B I think IM POSITIVE about A
GrogVix [38]3 years ago
3 0

Answer:

D. Wind

Explanation:

<em>Wind is Basically nothing how can you even use wind again when you cant even touch it you can only feel wind.</em>

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In which of the following locations is an island arc likely to form?
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<span>D. Along two converging oceanic plate boundaries</span>
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An object has a velocity of 8 m/s and a kinetic energy of 480 J. What is the mass of the object? (Formula: ) KE=1/2mv squared
Julli [10]

Answer:15kg

Explanation:

kinetic energy=ke=480J

Velocity=v=8m/s

Mass=m

Ke=0.5 x m x v x v

480=0.5 x m x 8 x 8

480=32m

Divide both sides by 32

480/32=32m/32

15=m

Mass=15kg

5 0
3 years ago
A sample of oxygen gas with a volume of 3.0 m3 is at 100 C. The gas is heated so that
vesna_86 [32]

\qquad\qquad\huge\underline{{\sf Answer}}

According to Charles law, if pressure remains constant, volume varies directly with temperature. so we can infer that :

\qquad\sf{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}

So, we can use this formula to find out the final temperature of the gas ~

Note : Take temperature in Kelvin ( 100°C = 373 K )

\qquad \sf  \dashrightarrow \:  \dfrac{3}{373}  =  \dfrac{6}{x}

\qquad \sf  \dashrightarrow \: x =  \dfrac{6}{3} \times 373

\qquad \sf  \dashrightarrow \: x = 2 \times 373

\qquad \sf  \dashrightarrow \: x =74 6 \: K

Now, convert it to Celsius ~

i.e 746 - 273 = 473° C

So, the final temperature of the gas will be equal to 473° C

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2 years ago
An astronaut is taking a space walk near the shuttle when her safety tether breaks. what should the astronaut do to get back to
GuDViN [60]

get hold of some of her equipment, and throw it away from the craft. she should recoil to the craft ... and  hope ???

7 0
3 years ago
5.00-kg particle starts from the origin at time zero. Its velocity as a function of time is given by v =6t^2 i + 2t j where v is
otez555 [7]

The concept of derivatives and integrals allows to find the results for the questions are the motion of a particle where the speed depends on time are:

       a)the position is:  r = 2 t³ i + t² j

       b) the position of the body on the y-axis is a parabola and on the x-axis it is a cubic function

       c) The acceleration is: a = 12 t i + 2 j

       d) the force is: F = 60 t i + 10 j

       e) the torque is:  τ = 40 t³ k^

       f) tha angular momentum is:  L = 4t³ - 6 t² k^

       g) The kinetic energy is: K = 2 m t² (9t² +1)

       h) The power is:   P = 2m (36 t³ + 2t)

Kinematics studies the movement of bodies, looking for relationships between position, speed and acceleration.

a) They indicate the function of speed.

        v = 6 t² i + 2t j

Ask the function of the position.   The velocity is defined by the variation of the position with respect to time

          v = \frac{dr}{dt}  

          dr = v dt

we substitute and integrate.

        ∫ dr = ∫ (6 t² i + 2t j) dt

        r - 0 = 6 \frac{t^3 }{3} \ \hat i + 2 \frac{t^2}{2 \ \hat j }  

       r = 2 t³ i + t² j

b) The position of the body on the y axis is a parabola and on the x axis it is a cubic function.

c) Acceleration is defined as the change in velocity with time.

           a = \frac{dv}{dt}  

           a = \frac{d}{dt} \ (6t^2 i + 2t j)  

           a = 12 t i + 2 j

d) Newton's second law states that force is proportional to mass times the body's acceleration.

          F = ma

          F = m (12 t i + 2 j)

          F = 5 12 t i + 2 j

          F = 60 t i + 10 j

e) Torque is the vector product of the force and the distance to the origin.

           τ = F x r

The easiest way to write these expressions is to solve for the determinant.

         \tau = \left[\begin{array}{ccc}i&j&k\\F_x&F_y&F_z\\x&y&z\end{array}\right]  

        \tau = \left[\begin{array}{ccc}i&j&k\\60t &10&0\\2t^3 &t^2&0\end{array}\right]  

       τ = (60t t² - 2t³ 10) k

       τ = 40 t³ k ^

f) Angular momentum

        L = r x p

        L =rx (mv)

        L = m (rxv)

The easiest way to write these expressions is to solve for the determinant.

       \left[\begin{array}{ccc}i&j&k\\2t^3 &t^2&0\\6t^2&2t&0\end{array}\right]  

    L = (4t³ - 6 t²) k

 

g) The kinetic energy is

            K = ½ m v²

            K = ½ m (6 t² i + 2t j) ²

            K = m 18 t⁴ + 2t²

            K = 2 m t² (9t² +1)

h) Power is work per unit time

           P = \frac{dW}{dt}dW / dt

The relationship between work and kinetic energy

           W = ΔK

     

          P = 2m \ \frac{d}{dt} ( 9 t^4 + t^2)

          p = 2m (36 t³ + 2t)

In conclusion with the concept of derivatives and integrals we can find the results for the questions are the motion of a particle where the speed depends on time are:

       a) The position is:  r = 2 t³ i + t² j

       b) The position of the body on the y-axis is a parabola and on the x-axis it is a cubic function

       c) The acceleration is: a = 12 t i + 2 j

       d) The force is: F = 60 t i + 10 j

       e) The torque is:  τ = 40 t³ k^

       f) The angular momentum is:  L = 4t³ - 6 t² k^

       g) The kinetic energy is: K = 2 m t² (9t² +1)

       h) The power is:   P = 2m (36 t³ + 2t)

Learn more here:  brainly.com/question/11298125

8 0
2 years ago
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