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nlexa [21]
3 years ago
10

Equations of motion ?​

Physics
1 answer:
kirill [66]3 years ago
4 0

Explanation:

There are five equations of motion:

v = at + v₀

Δx = v₀ t + ½ at²

Δx = ½ (v + v₀)t

v² = v₀² + 2aΔx

Δx = vt − ½ at²

Δx is the displacement

v₀ is the initial velocity

v is the final velocity

a is the acceleration

t is time

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A ball is thrown vertically upwards with a velocity of 9.8 m/s. Its velocity after 1 second will be
Reil [10]

\LARGE{ \underline{\underline{ \purple{ \bf{Required \: answer:}}}}}

GiveN:

  • Initial velocity = 9.8 m/s²
  • Accleration due to gravity = -9.8 m/s²
  • Time taken = 1 s

To FinD:

  • Final velocity of the ball?

Step-by-step Explanation:

Using the first Equation of motion,

⇒ v = u + gt

⇒ v = 9.8 + -9.8(1)

⇒ v = 0 m/s

The final velocity is hence <u>0</u><u> </u><u>m</u><u>/</u><u>s</u><u>.</u>

<h3>Note:</h3>

  • While solving questions of under gravity motions using equations of motion, remember the sign convection to avoid mistakes.
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5 0
2 years ago
what sense can people use to detect infrared radiation? we can see it as red light, we can feel it as heat, we can hear it as a
Ad libitum [116K]
We can feel it as heat, using the nerve endings in our skins.

7 0
3 years ago
Read 2 more answers
What is that distance traveled by an arrow going 78 m/s for 5 seconds
pshichka [43]

Explanation:

velocity=distance/time

distance= velocity×time

distance= 78×5=390m

8 0
2 years ago
The graph at the right shows the force needed to pull a bow back as the string is pulled further and further.
Sindrei [870]

A. 9 J

In a force-distance graph, the work done is equal to the area under the curve in the graph.

In this case, we need to extrapolate the value of the force when the distance is x=30 cm. We can easily do that by noticing that there is a direct proportionality between the force and the distance:

F=kx

where k is the slope of the line. We can find k, for instance chosing the point at x=5 cm and F=10 N:

k=\frac{F}{x}=\frac{10 N}{5 cm}=2 N/cm

And now we can calculate the work by calculating the area under the curve until x=30 cm, F=60 N:

W=\frac{1}{2} (height) (base)= \frac{1}{2}(60 N)(0.30 m)=9 J


B. 24.5 m/s

The mass of the arrow is m=30 g=0.03 kg. The kinetic energy of the arrow when it is released is equal to the work done by pulling back the bow for 30 cm:

W=K=\frac{1}{2}mv^2

where m is the mass of the arrow and v is its speed. By re-arranging the formula and using W=9 J, we find the speed:

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2\cdot 9J}{0.03 kg}}=24.5 m/s

8 0
3 years ago
A car starts from the origin and is driven 1.88 km south, then 9.05 km in a direction 47° north of east. Relative to the origin,
WINSTONCH [101]

Answer:

(a) θ = 55.85 degree

(b) 7.89 km

Explanation:

Using vector notations

A = 1.88 km south = 1.88 (- j) km = - 1.88 j km

B = 9.05 km 47 degree north of east

B = 9.05 ( Cos 47 i + Sin 47 j) km

B = (6.17 i + 6.62 j) km

Net displacement is

D = A + B

D = - 1.88 i + 6.17 i + 6.62 j = 4.29 i + 6.62 j

(a) Angle made with positive X axis

tanθ = 6.62 / 4.29 = 1.474

θ = 55.85 degree

(b) distance = Distance = \sqrt{(4.29)^{2} + (6.62)^{2}}

distance = 7.89 km

4 0
3 years ago
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