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OleMash [197]
3 years ago
9

A graph titled Position versus time for with horizontal axis time (seconds) and vertical axis position (meters). A straight blue

line runs with an upward slope from 0 seconds 3 meters to 4 seconds 15 meters.
The starting position of this object is
[___] m.

The object is traveling at a velocity of
[___] m/s.
Physics
2 answers:
koban [17]3 years ago
8 0

Answer:

The starting position of this object is

3m.

The object is traveling at a velocity of

3m/s.

professor190 [17]3 years ago
6 0

Answer:

3

3

Explanation:

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A person on a sled coasts down a hill and then goes over a slight rise with speed 2.7 m/s. The top of the rise can be modeled as
Paladinen [302]
Since we're dealing with radial acceleration around a circle, I used the radial acceleration equation a=v²/r. At the top of the hill, the force upward exerted by the hill is less than the weight of the sled.  if v is large enough the term (g-v²/r) will become 0 and the sled will fly off the ground as it reaches the peak.  Let me know if I can clarify any of my work.

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NEVERMIND ANSWER FOR SOME POINTS
saw5 [17]

Answer:

Answer what?

Explanation:

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3 years ago
Which of these is a process during which the hydrosphere interacts with the geosphere?
lana66690 [7]
<span>Answer: b. absorption of rainwater by soil

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4 years ago
Read 2 more answers
Two football players run towards each other along a straight path in Penrith Park in the clash between the Melbourne storms and
Agata [3.3K]

Answer:

a) v = 0.4799 m / s,  b)  K₀ = 1600.92 J,    K_f = 5.46 J

Explanation:

a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.

Initial instant. Before the crash

        p₀ = m v₁ + M v₂

where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive

Final moment. After the crash

       p_f = (m + M) v

as the system is isolated, the moment is preserved

       p₀ = p_f

       m v₁ + M v₂ = (m + M) v

       v =\frac{m v_1 + M v_2}{m+M}

let's calculate

        v = \frac{ -95 \ 3.75 \ + 111 \ 4.10}{95+111}

        v = 0.4799 m / s

b) let's find the initial kinetic energy of the system

         K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2

         K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2

         K₀ = 1600.92 J

the final kinetic energy

         K_f = ½ (m + M) v ^ 2

         k_f = ½ (95 + 111) 0.4799 ^ 2

         K_f = 5.46 J

3 0
3 years ago
An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc
Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

P=VI

Express this in terms of 'I'. This gives,

I=\frac{P}{V}

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, P_1=16\ W, V=110\ V

So, current drawn is given as:

I_1=\frac{16\ W}{110\ V}=0.145\ A

For incandescent bulb, P_2=100\ W, V=110\ V

So, current drawn is given as:

I_2=\frac{100\ W}{110\ V}=0.909\ A

For CFL bulb, P_3=24\ W, V=110\ V

So, current drawn is given as:

I_3=\frac{24\ W}{110\ V}=0.218\ A

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0
4 years ago
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