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Varvara68 [4.7K]
3 years ago
9

if an object is moving with constant speed then its distance time graph will be a straight line. a) along time axis b) along dis

tance axis c) parallel to time axis d)inclined to time axis.
Physics
1 answer:
Varvara68 [4.7K]3 years ago
5 0

Answer: I’m not sure but the time and distance will be increasing while they required the distance time so my answer is ((A))

Explanation:

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Pressure=force/area
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Suppose that you measure the light of a star between 640 and 680 nm and you get a very strong peak around 650 nm. what can you s
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Explanation:

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There is no difference between the question and hypothesis steps of scientific inquiry true or false
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A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
4 years ago
Anyone please help thank you
Thepotemich [5.8K]

Answer:

A) earth

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C) live

D) earth

Explanation:

hope i help

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3 years ago
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