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2 years ago

The graph shows two runners participating in a race.

Physics

2 answers:

Tatiana [17]2 years ago

8 0

Answer:

the guy above is correct its a

Explanation: just took the test

Olin [163]2 years ago

6 0

Answer:

'Daniela had a 5-meter head start, and Leonard caught up to her at 25 meters.'

Explanation:

hope that helps :)

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A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

$v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2$

$v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96$

Therefore, the velocity of tennis racket after collision is 14.96m/s

7 0

2 years ago

A mover loads a 100 kg box into the back of a moving truck by

NeX [460]

Answer:

2.7

Explanation:

The following data were obtained from the question:

Mass (m) of box = 100 Kg

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

Mechanical advantage of a ramp is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is given by:

Mechanical Advantage = Lenght / height

MA= L/H

With the above formula, we can obtain the mechanical advantage of the ramp as follow:

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

MA = 4/1.5

MA = 2.7

Therefore, the mechanical advantage of the ramp is 2.7

3 0

2 years ago

Since when was the light we see now emanating from the quasar? Note that the distance between the Earth and the quasar is 598 Mp

lakkis [162]

Hi hi hi hi hi hi hi hi hi hi hi

8 0

2 years ago

physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin

vova2212 [387]

Answer: Vb is the vector (-5.37m/s, 8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s, 8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

7 0

2 years ago

A 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10° with the horizontal. If there is no fricti

Alenkinab [10]

Answer:

$v_f = 3.45 m/s$

Explanation:

As we know that box was initially at rest

so here work done by all forces on the box = change in its kinetic energy

so we will have

$mg sin\theta L = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 10 kg

$\theta = 10^0$

L = 3.5 m

so we will have

$10(9.81)sin10 \times 3.5 = \frac{1}{2}(10)(v_f^2 - 0)$

$2(9.81) sin10 \times 3.5 = v_f^2$

so final speed is given as

$v_f = 3.45 m/s$

6 0

2 years ago