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NNADVOKAT
2 months ago
5

An object experiences an acceleration of -6.8 m/s​2.​ As a result, it accelerates from 54 m/s to a complete stop. How much dista

nce did it travel during that acceleration?
Physics
1 answer:
Maru [3.3K]2 months ago
3 0
The distance covered during the acceleration phase is d = 214.38 m. Given parameters include the acceleration of the object, a = -6.8 m/s², initial speed, u = 54 m/s, and final speed, v = 0. The equation for acceleration is defined as a = (v - u) / t. Therefore, rearranging gives t = (v - u) / a, resulting in t = (0 - 54) / (-6.8) = 7.94 s. The average speed of the object is V = (54 + 0)/2 = 27 m/s. The displacement is calculated as d = V x t = 27 x 7.94 = 214.38 m. Thus, the total distance traveled during that period of acceleration is 214.38 m.
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kicyunya [3294]
20.7 volts. The mass of an electron is 9.1 x 10⁻³¹ kg, and its wavelength is 0.27 x 10⁻⁹ m. The velocity of the electron can be determined using de Broglie's equation λ mv = h. Substituting the known values, we arrive at v = 2.7 x 10⁶ m/s. The potential difference through which the electron accelerates is noted, with the charge on an electron being 1.6 x 10⁻¹⁹ C. According to the conservation of energy, (0.5) mv² = q ΔV leads to ΔV = 20.7 volts.
4 0
2 months ago
A moving sidewalk 95 m in length carries passengers at a speed of 0.53 m/s. One passenger has a normal walking speed of 1.24 m/s
Maru [3345]
Answer: a) t = 1.8 x 10^2 seconds; b) t = 54 seconds; c) t = 49 seconds. Explanation: a) To determine the time of a stationary passenger on the sidewalk, we use the position formula. Given the constant speed of the walkway, we can calculate the time taken for set distances accordingly. This calculation extends into cases where combined velocities for walking are involved in subsequent queries.
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2 months ago
A student with a mass of 66.0 kg climbs a staircase in 44.0 s. If the distance between the base and the top of the staircase is
serg [3582]

power = 205.8 \: watt \\ solution \\ mass = 66 \: kg \\ time = 44 \: sec \\ distance = 14 \\ now \\ power = \frac{w}{t} \\ \: \: \: \: \: \: \: \: = \frac{f \times d}{t} \\ \: \: \: \: \: = \frac{m \times g \times d}{t} \\ \: \: \: \: \: \: = \frac{66 \times 9.8 \times 14}{44} \\ \: \: \: \: = \frac{9055.2}{44} \\ \: \: \: \: \: = 205.8 \: watt \\ hope \: it \: helps

4 0
2 months ago
A 2.0-kg projectile moves from its initial position to a point that is displaced 20 m horizontally and 15 m above its initial po
Yuliya22 [3333]

Answer:

W = 294 J

Explanation:

provided,

mass of the projectile = 2 Kg

horizontal displacement = 20 m

vertical displacement = 15 m

work performed by the gravitational force =?

the work done by gravitational force only accounts for vertical motion.

force due to gravity =  m g

= 2 x 9.8 = 19.6 N

work is equal to force x displacement

W = F x s

W = 19.6 x 15

W = 294 J

7 0
2 months ago
Read 2 more answers
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