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umka21
19 days ago
5

In an inertial frame of reference, a series of experiments is conducted. in each experiment, two or three forces are applied to

an object. the magnitudes of these forces are given. no other forces are acting on the object. in which cases may the object possibly remain at rest?
Physics
1 answer:
Yuliya22 [2.4K]19 days ago
8 0
Objects will stay in a stationary position if the total force acting on them amounts to zero; this occurs when equal forces are applied in opposite directions. According to Newton's second law, if the net force on an object is zero, it will not move.
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Compare the momentum of a 6,300-kg elephant walking 0.11 m/s and a 50-kg dolphin swimming 10.4 m/s. your answer
inna [2205]
<span>First, apply Newton's second law of motion: F = ma. Force equals mass times acceleration. This law describes force as the product of mass multiplied by acceleration (which is different from velocity). As acceleration is the variation in velocity over time, we have force = (mass * velocity) / time, leading us to conclude that (mass * velocity) / time will equal momentum / time. Hence, we derive the equation mass * velocity = momentum. Momentum = mass * velocity. For the elephant, with a mass of 6300 kg and velocity of 0.11 m/s, Momentum = 6300 * 0.11, resulting in P = 693 kg (m/s). For the dolphin, having a mass of 50 kg and moving at 10.4 m/s, Momentum = 50 * 10.4, yielding P = 520 kg (m/s). Thus, the elephant has a greater momentum (P) due to its larger size.</span>
5 0
1 month ago
A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/
ValentinkaMS [2425]

The formula for range is:

R = \frac{v_o^2 sin2\theta}{g}

Given values are:

v_0=37.2m/s

where θ equals 14.1 degrees

g=9.80m/s^2

Using the equation above,

R = \frac{37.2^2 sin2*14.1}{9.80}

The calculated range is 66.7 meters.

Therefore, the range is approximately 66.1 meters.

5 0
1 month ago
Read 2 more answers
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
serg [2593]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: The speed of a wave on a string under tension can be determined using the following:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} denotes tension (N)

μ refers to linear density (kg/m)

Calculating the velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

Distance a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5×10^{-6}  m.

When tension is doubled:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Distance in the same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With the increased tension, it moves \Delta x = 15.4×10^{-5} m

4 0
17 days ago
Astronomers determine that a certain square region in interstellar space has an area of approximately 2.4 \times 10^72.4×10 ​7 ​
Sav [2226]

Answer:

1.5 × 10³⁶ light-years

Explanation:

A particular square area in interstellar space measures roughly 2.4 × 10⁷² (light-years)². To find the area of a square, the following formula is utilized:

A = l²

where,

A represents the area of the square

l denotes the length of one side of the square

Thus, l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years

5 0
26 days ago
An airplane cruising at a constant velocity and altitude. Which of the following diagrams best represents the four forces of the
Yuliya22 [2420]

Answer:

Please include the diagrams and repost them.

7 0
21 day ago
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