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nika2105 [10]
3 years ago
9

1. calculate agni's average speed during the race.

Physics
1 answer:
MakcuM [25]3 years ago
7 0
The formula to find the speed/velocity is : 
v = Δx/Δt = (\frac{ x_{final} - x_{initial} }{ t_{travel} }

B.
1. From the graph given, we can get an information such as the distance travelled, and the time given.
2. If the graph are horizontal, it means that the Agni's speed is zero because Agni doesn't move.
3. If the question ask the velocity during all time, we can say that the velocity is :
.v =  \frac{x}{t} =  \frac{50 km}{2.5 hour} = 20 \frac{km}{h}
4. So, the final velocity can be found by looking the third part of the graph. So,..
v =  \frac{x}{t} =  \frac{(50-35) km }{0.5 hour} = 30 \frac{km}{h}

Hope this will help you :)
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Explanation:

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pls mark brainliest :P

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2 years ago
Use the component form of newton's second law to write an expression for the x component of the net force, σfx
nordsb [41]

Draw a free body diagram to show which forces act in the x and y directions. The x component equation is σfx = 0. The σfx being all the forces acting in the x direction.
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Which nucleus completes the following equation?<br> A. 299 Np<br> B. 20Pa<br> C. 2 Pa<br> D. - Np
blondinia [14]

Answer:

Option D. ²³⁹₉₃Np

Explanation:

Let the unknown be ʸₓA.

Thus, the equation becomes:

²³⁹₉₂U —> ⁰₋₁e + ʸₓA

Next, we shall determine the x, y and A. This can be obtained as follow:

92 = –1 + x

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92 + 1 = x

93 = x

x = 93

239 = 0 + y

239 = y

y = 239

ʸₓA => ²³⁹₉₃A => ²³⁹₉₃Np

Thus, the complete equation is:

²³⁹₉₂U —> ⁰₋₁e + ²³⁹₉₃Np

7 0
3 years ago
What do extremely massive stars end their life as?
IrinaVladis [17]

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6 0
3 years ago
Read 2 more answers
A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0
1 year ago
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