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lubasha [3.4K]
3 years ago
7

A projectile is fired directly upwards at 49.4 m/s. A second projectile is dropped from rest at some higher elevation at the ins

tant the first projectile is fired and passes the first projectile 3.00 s later. From the frame of reference of the first projectile, what is the velocity of the second projectile as it passes by
Physics
1 answer:
8090 [49]3 years ago
8 0

Answer:

Explanation:

velocity of first projectile after 3 s

v = u - gt

v = 49.4 - 9.8 x 3

= 20 m /s

Velocity of second projectile after 3 s after being dropped from rest

v = u + gt

= 0 + 9.8 x 3

= 29.4 m /s

They will be moving in opposite direction at the time of meeting , so their relative velocity

= 20 + 29.4 = 49.4 m /s

From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .

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What three forms of energy are represented when a match is lit?
AysviL [449]

Answer:

Three forms of energy when a match is being lit are potential, kinetic and thermal.

hope this helps! :)

Explanation:

4 0
3 years ago
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I'm working from a study guide and I got a question asking:
Harlamova29_29 [7]

you need to use hours to make it easy to convert to m/s. Change 15 minutes to hours

Note distance is in km so change the time to hours to get km/hour

1 hour = 60mins so 15 minutes will be 0.25 hours

10/0.25 = 40 km/ hour

To change that to m/second, change km to meters and hour to second

it will be (40x1000) meters /3600seconds

11.1111 m/s

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3 years ago
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El peso normal de un estudiante de secundaria es 725 N y el área de los dos zapatos que usa es de 412 cm2 . La presión medida qu
kotykmax [81]

Answer:

Presión = 175,97 N/m²

Explanation:

Dados los siguientes datos;

Peso del alumno (fuerza) = 725N

Área de zapatos = 412 cm² a metros cuadrados = 412/100 = 4.12 metros

Para encontrar la presión, usaríamos la siguiente fórmula;

Presión = fuerza / área

Presión = 725 / 4.12

Presión = 175,97 N/m²

8 0
3 years ago
Bone has a Young's modulus of about
Oksanka [162]

Answer:

505.62 mm

Explanation:

Young's modulus of a material is given by,

Y=\frac{Stress}{Strain}

Stress is the force experienced by the material per unit area (which is why stress has the same units as pressure).

Strain is the ratio between the change in length/area/volume to the original length/area/volume of the material (and hence a dimensionless quantity). Basically, it gives us a measure of the change in dimensions due to the stress experienced.

So, for the given problem,

Y=\frac{Stress}{\Delta L/L}

or, 1.8\times100=\frac{1.59\times108}{\Delta L/0.53}      (since we need to find the compression of the bone at the critical stress point)

or, \Delta L=\frac{1.59\times108\times0.53}{1.8\times100} m=0.50562m=505.62mm

8 0
3 years ago
Two point charges are on the y axis. A 3.90-µC charge is located at y = 1.25 cm, and a -2.4-µC charge is located at y = −1.80 cm
ladessa [460]

Answer:

a) 1.6*10^6 V

b) 13.35*10^6 V

Explanation:

The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:

V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}    (1)

q1 = 3.90µC = 3.90*10^-6 C

q2 = -2.4µC = -2.4*10^-6 C

r1 = 1.25 cm = 0.0125 m

r2 = -1.80 cm = -0.018 m

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

You replace all the parameters in the equation (1):

V=k[\frac{q_1}{r_1}+\frac{q_2}{r_2}]\\\\V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0125m}+\frac{-2.4*10^{-6}C}{0.018m}]=1.6*10^6V

hence, the total electric potential is approximately 1.6*10^6 V

b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:

r1 = 0.0150m - 0.0125m = 0.0025m

r2= 0.015m + 0.018m = 0.033m

Then, you replace in the equation (1):

V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0025m}+\frac{-2.4*10^{-6}C}{0.033m}]=13.35*10^6V

hence, for y = 1.50cm you obtain V = 13.35*10^6 V

4 0
3 years ago
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