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Vinil7 [7]
3 years ago
14

An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0

m from time t = 1.00 s to time t = 2.00 s. What is the acceleration of the object?
Physics
1 answer:
Reptile [31]3 years ago
7 0

Answer:

28 m/s^2

Explanation:

distance, s = 14 m

time, t = 2 - 1 = 1 s

initial velocity, u = 0 m/s

Let a be the acceleration.

Use third equation of motion

s = ut + \frac{1}{2}at^{2}

14 = 0 + \frac{1}{2}a\times 1^{2}

a = 28 m/s^2

Thus, the acceleration is 28 m/s^2.

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A kite is hovering over the ground at the end of a straight 38-m line. the tension in the line has a magnitude of 17 n. wind blo
postnew [5]

Let the tension in the thread will be at angle theta

now by force balance in  x and y directions we can say

17 cos\theta = 18 cos60

17 cos\theta = 9

\theta = 58.03 degree

now for the height of the kite we can use

h = L sin\theta

h = 38 sin58.03

h = 32.23 m

<em>so it will be at height 32.23 m from ground</em>

3 0
3 years ago
Help yet again :) A hockey player is skating on the ice at 15km/h. He shoots the puck at 138 km/h according to a radar gun on th
IceJOKER [234]

Answer:

speed of puck acc. to the radar gun = 138 km/h

speed of player = 15 km/h

since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,

speed of puck = speed of player + speed of puck acc. to player

138 = 15 + speed of puck acc. to player

speed of puck acc. to player = 138 -15

speed of puck acc. to player = 123 km/h

Brainly this answer if you think it deserves it

7 0
3 years ago
Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo
yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour                    

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is   = 102^{\circ}-65.3^{\circ}=36.7^{\circ}

Speed of first plane  = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, r^2 representing the distance between the planes, we see that:

r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295

r = 741.6959 m

3 0
3 years ago
In a cyclic process, a gas performs 123 J of work on its surroundings per cycle. What amount of heat, if any, transfers into or
Margaret [11]

Answer:

123 J transfer into the gas

Explanation:

Here we know that 123 J work is done by the gas on its surrounding

So here gas is doing work against external forces

Now for cyclic process we know that

\Delta U = 0

so from 1st law of thermodynamics we have

dQ = W + \Delta U

dQ = W

so work done is same as the heat supplied to the system

So correct answer is

123 J transfer into the gas

8 0
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Answer:

42 I THE CORRECT ANSWER

Explanation:

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