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elena-s [515]
3 years ago
14

#1 “A persons body continuing to move forward even though the car comes to a sudden stop” Which newtons Law Do they pertain? A:

First Law B: Second Law C: Third Law
#2 “A fighter Pilot Feels Massive Amounts of forcé when his plane turns sharply” Which Newton Law?

#3 “ A Paddle wheel boat pushed on water and the water pushes back causing the boat to move” Which Netwons Law?
Physics
1 answer:
Aloiza [94]3 years ago
3 0
<span>#1 “A persons body continuing to move forward even though the car comes to a sudden stop” Which newtons Law Do they pertain?
A: First Law </span>

<span>#2 “A fighter Pilot Feels Massive Amounts of forcé when his plane turns sharply” Which Newton Law?
A: First Law.
</span><span><span>You were following the Newton's first law and kept your velocity straight until you departed from linear motion when you turn sharply; you are forced to follow the curve. <span>The force that the jet exerts on you is called centripetal force and is suitable for the center of curvature of the forced traveling path.</span></span>

</span><span>#3 “ A Paddle wheel boat pushed on water and the water pushes back causing the boat to move” Which Netwons Law?
C: Third Law</span>


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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i
NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 - 14.7t - 4.9t^2

4.9t^2 + 14.7t - 19.6 = 0

t^2 + 3t - 4 = 0

(t + 4)(t - 1) = 0

(t - 1) = 0

\boxed {t = 1 ~ second}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 + 14.7t - 4.9t^2

4.9t^2 - 14.7t - 19.6 = 0

t^2 - 3t - 4 = 0

(t - 4)(t + 1) = 0

(t - 4) = 0

\boxed {t = 4 ~ seconds}

The difference in the two ball's time in the air is:

\Delta t = 4 ~ seconds - 1 ~ second

\large {\boxed {\Delta t = 3 ~ seconds} }

<h2>Question B:</h2><h3>First Ball</h3>

v^2 = u^2 - 2gH

v^2 = (-14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

<h3>Second Ball</h3>

v^2 = u^2 - 2gH

v^2 = (14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 11.025 ~ m}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 25.725 ~ m}

The difference in the two ball's height after 0.500 s is:

\Delta h = 25.725 ~ m - 11.025 ~ m

\large {\boxed {\Delta h = 14.7 ~ m} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0
3 years ago
8. A car moving at 35 m/s has 675 joules of KE. What is the mass of the car?
andreyandreev [35.5K]
Kinetic energy is the energy possessed by a body in motion while potential energy is the energy of a body at rest.
Kinetic energy is given by E=1/2MV² where M is the mass of the body while V is the velocity of the body. 
To get mass we can use the formula M= 2 Ek/V² (Making M the subject)
hence mass  = (2 ×675)÷35²
                     = 1.102 kg 
6 0
3 years ago
A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west
ivolga24 [154]

Answer:

Magnitude of avg velocity, |v_{avg}| = 18.9 km/h

\theta' = 56.85^{\circ}

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = \frac{27}{60} h

Angle, \theta = 50^{\circ}

Time taken in 50^{\circ}east of due North direction, t' = 29 min =  \frac{29}{60} h

Time taken in west direction, t'' = 37 min =  \frac{27}{60} h

Solution:

Now, the displacement, 's' in east direction is given by:

\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km

Displacement in  50^{\circ} east of due North for 29.0 min is given by:

\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}

\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}

\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km

Now, displacement in the west direction for 37 min:

\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km

Now, the overall displacement,

\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}

\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}

\vec{s_{net}} =  16.03\hat{i} + 24.54\hat{j} km

(a) Now, average velocity, v_{avg} is given:

v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}

v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}

v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h

Magnitude of avg velocity is given by:

|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h

(b) angle can be calculated as:

tan\theta' = \frac{15.83}{10.34}

\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}

6 0
3 years ago
Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

7 0
3 years ago
Describe two reasons why scientists are looking for alternatives to fossil fuels
Evgesh-ka [11]
1) It is because they produce a lot of pollution in the atmosphere
2) They are limited in number and can't be recycled

Hope this helps!
3 0
3 years ago
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