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Anna35 [415]
2 years ago
7

What formula allows you to calculate the x component of a projectile?

Physics
2 answers:
Kay [80]2 years ago
8 0

X-component of a projectile in flight =

(initial x-component)

plus

(initial horizontal component of velocity) x (flight time so far)

RideAnS [48]2 years ago
4 0

Answer:

Explanation:

X- component = Horizontal component

Initial velocity = U

X- component =Ucos x angle

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Isabella drops a pen off her balcony by accident while celebrating the successful completion of a physics problem.
Studentka2010 [4]

Answer:

2 seconds

Explanation:

v = at + v₀

19.62 m/s = (9.81 m/s²) t + 0 m/s

t = 2 s

6 0
3 years ago
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A school bus takes 0.530 h to reach the school from your house. If the average velocity of the bus is 19.0 km/h to the east, wha
kifflom [539]

Answer:

x_{distance}=10.07km

Explanation:

Given data

time=0.530 h

Average velocity Vavg=19.0 km/s

To find

Displacement Δx

Solution

The Formula for average velocity is given as

V_{avg}=(x_{distance} )/(t_{time} )\\ x_{distance}=V_{avg}*(t_{time} )\\x_{distance}=(19.0km/h)*(0.530h)\\x_{distance}=10.07km

3 0
3 years ago
A particle makes 800 revolution in 4 minutes of a circle of 5cm. Find
vladimir1956 [14]

Answer:

i) The period of the particle is 0.3 seconds

ii) The angular velocity is approximately 20.94 rad/s

iii) The linear velocity is approximately 1.047 m/s

iv) The centripetal acceleration is approximately 6.98 m/s²

Explanation:

The given parameters are;

The number of revolution of the particle, n = 800 revolution

The time it takes the particle to make 800 revolutions = 4 minutes

The dimension of the circle = 5 cm = 0.05 m

Given that the dimension of the circle is the radius of the circle, we have;

i) The period of the particle, T = The time to complete one revolution

T = 1/(The number of revolutions per second)

∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s

The period, T = 3/10 seconds = 0.3 seconds

ii) The angular velocity, ω = Angle covered/(Time)

800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes

∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3

The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s

iii) The linear velocity, v = r × ω

∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s

iv) The centripetal acceleration, a_c = v²/r

∴ The centripetal acceleration, a_c = (π/3)²/(0.05) = 20·π/9

The centripetal acceleration, a_c = 20·π/9 m/s² ≈ 6.98 m/s²

4 0
2 years ago
25.5 m/s in 5.75 s. What is the acceleration?
anzhelika [568]
Acceleration units are in m/s^2 , so you take 25.5 m/s divided by 5.75s. Acceleration is the rate of change of velocity :)

follow me on instagram : imrajsingh :):) thanks
7 0
3 years ago
A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t
emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =F^> × x^>

W = Fxcos\theta    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = Fxcos( 0° )

W = Fx ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

W_{net = ΔKE

= \frac{1}{2}mv² -  \frac{1}{2}mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

Fx = \frac{1}{2}mv² -  \frac{1}{2}mu²

we make F, the subject of formula

F = \frac{m}{2x}( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = \frac{830}{(2)(0.255)}( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

6 0
2 years ago
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