PHYSICS COLLEGE

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.60 m/s^2 . At the same instant a truck, traveling with a constant speed of 23.5 m/s , overtakes and passes the car. a. How far beyond its starting point does the car overtake the truck?b. How fast is the car traveling when it overtakes the truck?

Answers

Answer 1

Answer:

Answer:

306.8264448 m

47.0016 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled by car

$s_c=ut+(1)/(2)at^2\n\Rightarrow s_c=(1)/(2)at^2$

Distance traveled by truck

$s_t=ut$

In order to overtake both distances should be equal

$(1)/(2)at^2=ut\n\Rightarrow (1)/(2)at=u\n\Rightarrow t=(2u)/(a)\n\Rightarrow t=(2* 23.5)/(3.6)\n\Rightarrow t=13.056\ s$

$s_c=(1)/(2)at^2\n\Rightarrow s_c=(1)/(2)3.6* 13.056^2\n\Rightarrow s_c=306.8264448\ m$

The distance the car has to travel is 306.8264448 m

$v^2-u^2=2as\n\Rightarrow v=√(2as+u^2)\n\Rightarrow v=√(2* 3.6* 306.8264448+0^2)\n\Rightarrow v=47.0016\ m/s$

The speed of the car when it overtakes the truck is 47.0016 m/s

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Which of the following is correct? *PLEASE HELP MEEEE
1 cm = 100 m
1 mm = 100 cm
100 mm = 1 cm
1 m = 100 cm

Answers

Answer:

The last one

1m = 100 cm

Explanation:

If you do not trust me look it up

Two objects of equal mass are a distance of 5.0 m apart and attract each other with a gravitational force of 3.0 x 10^-7 N find their mass.A) 150 kg
B) 9.8 kg
C) 11.000 kg
D) 340 kg

Answers

Answer

I Think Its 150

*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is the
acceleration of the ball?

Answers

Answer:

-54,200 m/s^2

Explanation:

a=(vf-vi)/t

NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider such a spacecraft that has a diameter of d = 148 m. What is the speed v, in meters per second, the spacecraft must rotate at its outer edge to replicate the force of gravity on earth?

Answers

The speed of the spacecraft at its outer edge is 26.93 m/s.

The given parameters;

diameter of the spacecraft, d = 148 m
radius of the spacecraft, r = 74 m

The speed of the spacecraft at its outer edge is calculated as follows;

$F_g = F_c\n\nmg = (mv^2)/(r) \n\nv^2 = rg\n\nv = √(rg) \n\nv = √((74)(9.8)) \n\nv = 26.93 \ m/s$

Thus, the speed of the spacecraft at its outer edge is 26.93 m/s.

Learn more here:brainly.com/question/20905151

Answer:

Explanation:

Given

diameter of spacecraft $d=148\ m$

radius $r=74\ m$

Force of gravity $F_g$ =mg

where m =mass of object

g=acceleration due to gravity on earth

Suppose v is the speed at which spacecraft is rotating so a net centripetal acceleration is acting on spacecraft which is given by

$F_c=(mv^2)/(r)$

$F_c=F_g$

$(mv^2)/(r)=mg$

$(v^2)/(r)=g$

$v=√(gr)$

$v=√(1450.4)$

$v=38.08\ m/s$

Which of these 23rd chromosomecombinations is likeliest to result in a
person with male and female traits?
ΧΟ
XXX
XXY
XY

Answers

Sorry if I’m wrong but I think it’s XO since o is not a sex chromosome

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).1. For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.Express your answer in terms of g and the variables m, l, x, and θ.(U^G=?)2. Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)Express your answer in terms of g and the variables m, l, and θ.

Answers

Answer:

UG (x) = m*g*x*sin(Q)

Vx,f (x)= sqrt (2*g*x*sin(Q))

Explanation:

Given:

- The length of the friction less surface L

- The angle Q is made with horizontal

- UG ( x = L ) = 0

- UK ( x = 0) = 0

Find:

derive an expression for the potential energy of the block-Earth system as a function of x.

determine the speed of the block at the bottom of the incline.

Solution:

- We know that the gravitational potential of an object relative to datum is given by:

UG = m*g*y

Where,

m is the mass of the object

g is the gravitational acceleration constant

y is the vertical distance from datum to the current position.

- We will consider a right angle triangle with hypotenuse x and angle Q with the base and y as the height. The relation between each variable can be given according to Pythagoras theorem as follows:

y = x*sin(Q)

- Substitute the above relationship in the expression for UG as follows:

UG = m*g*x*sin(Q)

- To formulate an expression of velocity at the bottom we can use an energy balance or law of conservation of energy on the block:

UG = UK

- Where UK is kinetic energy given by:

UK = 0.5*m*Vx,f^2

Where Vx,f is the final velocity of the object @ x:

m*g*x*sin(Q) = 0.5*m*Vx,f^2

-Simplify and solve for Vx,f:

Vx,f^2 = 2*g*x*sin(Q)

Hence, Velocity is given by: