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Advocard [28]
3 years ago
5

A sprinter accelerates from rest to a top speed of 24 km/h in 2 seconds and then runs at a constant velocity for the rest of the

100 meter dash,
a- what was the distance he covered in the first 2 seconds

b- what was his time for the whole race
Physics
1 answer:
Tems11 [23]3 years ago
5 0

Answer:

a. 6.67 meters

b. 16 seconds

Explanation:

a. v=(24×1000)/3600=6.67 m/s

v=u-at

a=(v-u)/t=(6.67-0)/2=3.335 m/s^2

s=ut+(1/2) at^2 = 0×2+.5×3.335×(2)²=6.67 meters

b. After 2seconds, he ran at a constant velocity. From a, in 2 seconds, he crossed 6.67 meters.

So, distance left =(100-6.67)=93.33 meters

to cross 93.33 meters,he needed,

s=vt

93.33=6.67t

t=13.99≡14 seconds

So, Total (93.33+6.67)=100 meters were covered in (14+2)=16 seconds

Hope, this helps you.

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A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.75 ms2.75 ms from an init
kumpel [21]

Answer:

2677.3 N

Explanation:

v₀ = initial speed of the hand = 4.75 m/s

v  = final speed of the hand = 0 m/s

m = Total mass of hand and forearm = 1.55 kg

t = time interval for hand to come to rest = 2.75 ms = 0.00275 s

F = Force applied on the leg

Using Impulse-change in momentum equation

F t = m (v - v₀)

F (0.00275) = (1.55) (0 - 4.75)

F = - 2677.3 N

magnitude of force = 2677.3 N

6 0
3 years ago
 why are waves faster in liquids and gases than in air
Alika [10]

Answer:

Sound travels faster in liquids than in gases because molecules are packed more closely together. This means that when the water molecules begin to vibrate, they quickly begin to collide with each other forming a rapidly moving compression wave. Sound travels over four times faster than in air

Explanation:

4 0
3 years ago
Read 2 more answers
While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration=2.40m/s^{2}, c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = 2.40m/s^{2}

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}

after cancelling the pertinent units, I get that  my answer will be given in meters. So I get:

x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}

which solves to:

x=172.8m

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

a=\frac{V_{f}-V_{0} }{t}

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

at=V_{f}-V_{0}

and finally I can add V_{0} to both sides so I get:

V_{f}=at+V_{0}

and now I can proceed and substitute the known values:

V_{f}=at+V_{0}

V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0

which solves to:

V_{f}=28.8m/s

8 0
3 years ago
Read 2 more answers
Help me yall it due in a few minutes :((()
vaieri [72.5K]

Answer:

B. blocks 2 & 3.

Explanation:

Block 1 has equal & opposite forces acting on it.

Block 2 has 5N on one side, 3N on the other. It will move in the direction the 5N of force is pushing.

Block 3 has no opposing force.

6 0
3 years ago
A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar
m_a_m_a [10]

Answer:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

#Where u is in meters and t in seconds.

Explanation:

Given that :m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s

From \omega=kL we have:

k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2

From F_d(t)=-\gamma u\prime(t) we have that:

\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m

Now,given that the initial value problem is given by:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

Hence,the position of u at time t is given by

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\, u in meters,t in seconds.

4 0
3 years ago
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