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jekas [21]
3 years ago
11

A car starts from rest and travels for t1 seconds with a uniform acceleration a1. The driver then applies the brakes, causing a

uniform acceleration a2. If the brakes are applied for t2 seconds, determine the following. Answers are in terms of the variables a1, a2, t1, and t2. (a) How fast is the car going just before the beginning of the braking period?
Physics
1 answer:
Triss [41]3 years ago
5 0

Explanation:

It is given that,

Initially the car is at rest and travels for t₁ seconds with a uniform acceleration a₁. The driver then applies the brakes, causing a uniform acceleration a₂, If the brakes are applied for t₂ seconds.

We need to find the speed of the car just before the beginning of the braking period.

Using the formula of acceleration. It is given by :

a_1=\dfrac{v-u}{t_1}

u = 0

v=a_1\times t_1

So, just before the beginning of the braking period the speed of the car is a_1\times t_1. Hence, this is the required solution.

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A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ po
juin [17]

Answer:

Following are the solution to this question:

Explanation:

Law:

\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}

      = 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}

\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}

\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\

4 0
3 years ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
3 years ago
A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th
inn [45]

Answer:

|\vec r|=339.82\ m

\theta=-6.67^o

Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

\vec r=\vec r_1+\vec r_2

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

x=rcos\theta

y=rsin\theta

And the vector is expressed as

\vec z==

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

\vec r_1==\ km=\ m

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

\vec r_2==\ m

The total displacement is

\vec r=\ m+\ m

\vec r=\ m

In (magnitude,angle) form:

|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m

\boxed{|\vec r|=339.82\ m}

\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169

\boxed{\theta=-6.67^o}

5 0
3 years ago
A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res
Luden [163]
The first thing you should know for this case is the definition of distance.
 d = v * t
 Where,
 v = speed
 t = time
 We have then:
 d = v * t
 d = 9 * 12 = 108 m
 The kinetic energy is:
 K = ½mv²
 Where,
 m: mass
 v: speed
 K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
 The work due to friction is
 w = F * d
 Where,
 F = Force
 d = distance:
 w = 400 * 108 = 4.32 * 10 ^ 4
 The power will be:
 P = (K + work) / t
 Where,
 t: time
 P = 2.86 * 10 ^ 5/12 = 23.9 kW
 answer:
 the average power developed by the engine is 23.9 kW
8 0
3 years ago
Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
Evgen [1.6K]

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.

4 0
3 years ago
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