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I am Lyosha [343]
3 years ago
14

the car starts from a stop to travel 1100 meters in 14 seconds. it is clocked at 65 m/s at point k. find its average speed and a

cceleration ​
Physics
1 answer:
inysia [295]3 years ago
4 0

Answer:

The average velocity of the car is, V = 74.04 m/s

Explanation:

Given data,

The initial velocity of the car, u = 0 m/s

The displacement of the ca, S = 1100 m

The time period of travel, t = 14 s

The velocity of the car at point k, v = 65 m/s

Using the II equation of motion,

                      S = ut + ½  at²

Substituting the given values,

                      1100 = 0 + ½ x a x 14²

                          a = 11.22 m/s²

Using the III equation of motion

                         v² = u² + 2 as

                          v = √(2as)              (∵ u = 0)

Substituting,

                           v = √(2 x 11.22 x 1100)

                              = 157.11 m/s

The average speed of the car,

                        V=\frac{0+65+157.11}{3}

                        V = 74.04 m/s

Hence, the average velocity of the car is, V = 74.04 m/s

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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
2 years ago
Question 7 (1 point)
ioda

Answer:

1150 secs

distance = speed x time

time = distance / speed

230,000 / 200 = 1150

time = 1150 seconds

7 0
3 years ago
a 1 gram spiders sits on a platform rotating at 78 rpm. the spider is 15 cm from the centre disk. find the speed of the spider
olga nikolaevna [1]

The spider is traveling in a circle with radius = 15cm

The circumference of any circle = <em>2 pi (radius)</em>
The circumference of the spider's path = 2 pi (15 cm) = 30 pi cm

The spider completes a trip around this path 78 times per minute.
Its speed, relative to you, is   

                               (78) x (30 pi) cm/min =

                                       2,340 pi cm/min =  7,351.33 cm/min =

                                     <em>  73.5133 meter/min =</em>

                                       <em>4.411 km/hr =</em>

                                         <em>2.74  miles/hour

</em>
(After the last appearance of pi,
all numbers are rounded.)<em>

</em>
8 0
3 years ago
A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg
Natali5045456 [20]

Answer: Vf = 2,400,000 m/s

Explanation:

1) The only relevant force is the electrostatic force

2) The formula for the electrostatic force is F = E×q

Where E is the electric field and q is the magnitude of the charge.

3) Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magntitude of the electric forces acting in both proton and electron are the same

4) Fe = Fp (Fe stands for force on the electron and Fp stands for force on the proton).

5) Using second law of Newton, Force = mass × acceleration

Fe = Me × Ae (Me is mass of the electron, and Ae is acceleration of the electron)

Fp = Mp × Ap (Mp is mass of the proton, and Ap is acceleration of the proton)

⇒Me × Ae = Mp × Ap

⇒ Ae = Mp × Ap / Me

6) Now, state the equations for the velocity in uniformly accelerated motion:

i) Vf² = Vo² + 2ad

Vo² = 0 for both cases, and d is the same distance.

⇒ Vf² = 2ad

ii) For the proton Vf² = 2(Ap)(d) ⇒ Ap = Vf² / (2d)

⇒ Ap = (55,000 m/s)² / (2d)

iii) For the electron Vf² = 2(Ae)² (2d)

iv) Using Ae = Mp × Ap / Me (found prevously):

Vf² = Mp × (55,000 m/s)² / (2d) × (2d) / Me

⇒ Vf² = Mp × (55,000 m/s)² / Me

Taking square root in both sides:

⇒ Vf = 55,000 m/s × √ [Mp / Me]

7) These are the values for the masses of a proton and an electron:

Mp = 1.67 × 10⁻²⁷ kg

Me = 9.11×10⁻³¹ kg

8) Replace and compute:

Vf = 55,000 m/s × √ [ 1.67 × 10⁻²⁷ kg / 9.11×10⁻³¹ kg] = 2,354,841.8 m/s

Round to two significan digits: Vf = 2,400,000 m/s

5 0
3 years ago
Read 2 more answers
A child and sled with a combined mass of 50 kg, start from rest and slide down a frictionless hill that is 7.5 meters high. what
amid [387]
The mechanical enegia is the sum of the kinetic energy plus the potential energy
 Kinetic energy = (1/2) * m * v ^ 2
 Potential energy = m * g * h
 Mechanical energy = (1/2) * m * v ^ 2 + m * g * h
 What is the mechanical energy of the sled at the top? 
 Mechanical energy = (1/2) * m * v ^ 2 + m * g * h
 Mechanical energy = (1/2) * (50) * (0) ^ 2 + (50) * (9.8) * (7.5) = 3675
 Mechanical energy = 3675J
 What is the mechanical energy of the sled at the bottom? 
 By conservation of energy we have that the energy in point 1 is equal to the energy in point 2
 Mechanical energy = 3675J
 What is the speed of the sled at the bottom of the hill?
 Mechanical energy = 3675J = (1/2) * m * v ^ 2
 clearing up v we have
 (1/2) * (50) * v ^ 2 = 3675
 v ^ 2 = 3675 * (2) * (1/50)
 v = root (3675 * (2) * (1/50)) = 12.12 m / s
 answer
 3675J
 3675J
 12.12 m / s
3 0
3 years ago
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