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jek_recluse [69]
2 years ago
13

A person driving a car suddenly applies the brakes. The car takes 4 s to come to rest while traveling 20 m at constant accelerat

ion. Can the speed of the car immediately before the brakes were applied be determined without first determining the car’s acceleration?
Physics
1 answer:
Zinaida [17]2 years ago
7 0

Answer:

Yes we can find the initial velocity of car without finding acceleration.

u = 10 m/s.

Explanation:

Given that

s=20 m

Car takes 4 s to come in rest.

We know that when acceleration is constant then we can apply motion equation

v=u+at        ----------1

s=ut+\dfrac{1}{2}at^2       ------2

From equation 1 and 2

s=ut+\dfrac{1}{2}t^2\left (\dfrac{v-u}{t} \right )

So we can say that

s= \left(\dfrac{v+u}{2}\right)t

Given that the velocity of car at final condition will be zero (v=0)

s= \left(\dfrac{0+u}{2}\right)t

s= \left(\dfrac{u}{2}\right)t

From the above equation we can find the initial velocity of car without finding the acceleration

20= \dfrac{u}{2}\times 4

u = 10 m/s

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The Orion nebula is one of the brightest diffuse nebulae in the sky (look for it in the winter, just below the three bright star
Oxana [17]

Answer:

T=183.21K

Explanation:

We have to take into account that the system is a ideal gas. Hence, we have the expression

PV=nRT

where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the ideal gas constant.

Thus, it is necessary to calculate n and V

V is the volume of a sphere

V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5.9*10^{15}m)^3=8.602*10^{47}m^3

V=8.86*10^{50}L

and for n

n=\frac{(4000M_s)/(2*mH)}{6.022*10^{23}mol^{-1}}=3.95*10^{36}mol

Hence, we have (1 Pa = 9.85*10^{-9}atm)

T=\frac{PV}{nR}=\frac{(6.8*10^{-9}*9.85*10^{-6}atm)(8.86*10^{50}L)}{(0.0820\frac{atm*L}{mol*K})(3.95*10^{36}mol)}\\\\T=183.21K

hope this helps!!

8 0
2 years ago
The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction betwe
Novosadov [1.4K]

Answer:

a) a= 8.33 m/s²,    T = 12.495 N , b)    a = 2.45 m / s²

Explanation:

a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.

We apply Newton's second law to the lower charge

            fr₁ + fr₂ = ma

The equation for the force of friction is

          fr = μ N

Y Axis

         N - W₁ –W₂ = 0

         N = W₁ + W₂

         N = (m₁ + m₂) g

Since the beams are the same, it has the same mass

        N = 2 m g

We replace

           μ₁ 2mg + μ₂ mg = m a

          a = (2μ₁ + μ₂) g

          a = (2 0.30 + 0.25) 9.8

          a= 8.33 m/s²

Let's look for cable tension with beam 2

          T = m₂ a

          T = 1500 8.33

          T = 12.495 N

b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal

           fr = m₂ a₂

           N- w₂ = 0

          N = W₂ = mg

          μ₂ mg = m a₂

          a = μ₂ g

          a = 0.25 9.8

          a = 2.45 m / s²

4 0
2 years ago
Problem: An ice hockey player hits a puck of mass 0.15 kilograms with a force of 100 newtons in the horizontal direction. What i
valina [46]
The acceleration produced in a body is always in the direction of the resultant force acting on the body. Therefore, we may determine the horizontal acceleration using the horizontal force applied. To do this, we may apply the mathematical form of Newton's second law:

Force = mass * acceleration
acceleration = force / mass

Substituting the values,
a = 100 / 0.15
a = 666.7 m/s²

The acceleration of the hockey puck is 670 m/s²
7 0
2 years ago
Read 2 more answers
Two cyclists start on a race between points A and D on two different routes. Cyclist X takes the route passing through the equid
ollegr [7]
The displacement is the shortest distance between two points, which is 546.41. The displacement for both is 546.41 meters

Average velocity of X = (200 + 200 + 200) / 30
Average velocity of X = 20 m/s

Average velocity of Y = 546.41 / 30 = 18.2 m/s
8 0
2 years ago
Read 2 more answers
You are target shooting using a toy gun that fires a small ball at a speed of 15 m/s. When the gun is fired at an angle of 31 de
Ray Of Light [21]

Answer: 13.1° from the horizontal

Explanation: For projectile

Horizontal distance R of a projectile is:

R= U×Usin2x/g

Where U is the speed of projectile, x is angle of projectile and g= acceleration due to gravity=9.8m/s

R= 15×15sin(2×31)/9.8= 225sin(62)/9.8=20.27m

Therefore if R is halved.

R/2 = 20.27/2=10.14m

Hence the angle for this case is.

Making sin(2x) the subject of formula

Sin2x= Rg/U×U

R is now 10.14 in this case.

Sin2x= 10.14×9.8/15×15=0.4415

2x=arcsin0.4415=26.2

x= 26.2/2

x= 13.1°

Good luck...

3 0
2 years ago
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