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3 years ago

What is projectile motion caused by?

2 answers:

6 0

Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.

lora16 [44]3 years ago

4 0

Answer:

A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.

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Aman tosses a jart upward with a velocity of 14.1 m/s a 60° angle

crimeas [40]

Answer:

hold on let me check my answer

Explanation:

8 0

4 years ago

If atmospheric pressure suddenly changes from 1.00 atm to 0.896 atm at 298 k, how much oxygen will be released from 3.30 l of wa

xxMikexx [17]

At a temperature of 298 K, the Henry's law constant is 0.00130 M/atm for oxygen. The solubility of oxygen in water 1.00 atm would be calculated as follows:

S = (H) (Pgas) = 0.00130 M / atm x 0.21 atm = 0.000273 M

At 0.890 atm,
S = (H)(Pgas) = 0.00130 M / atm x 0.1869 atm = 0.00024297 M

If atmospheric pressure would suddenly change from 1.00 atm to 0.890 atm at the same temperature, the amount of oxygen that will be released from 3.30 L of water in an unsealed container would be as follows

3.30 L x (0.000273 mol / L) = 0.0012012 mol

3.30 L x (0.00024297 mol / L) = 0.001069068 mol

0.0012012 mol - 0.001069068 mol = 0.000132 mol

6 0

4 years ago

Force X has a magnitude of 1260 pounds, and Force Y has a magnitude of 1530 pounds. They act on a single point at an angle of 4

weeeeeb [17]

Answer:

Fe= 2579.68 P

α= 24.8°

Explanation:

Look at the attached graphic

we take the forces acting on the x-y plane and applied at the origin of coordinates

FX = 1260 P , horizontal (-x)

FY = 1530 P , forming 45° with positive x axis

x-y components FY

FYx= - 1530*cos(45)° = - 1081.87 P

FYy= - 1530*sin(45)° = - 1081.87 P

Calculation of the components of net force (Fn)

Fnx= FX + FYx

Fnx= -1260 P -1081.87 P

Fnx= -2341.87 P

Fny=FYy

Fny= -1081.87 P

Calculation of the components of equilibrant force (Fe)

the x-y components of the equilibrant force are equal in magnitude but in the opposite direction to the net force components:

Fnx= -2341.87 P, then, Fex= +2341.87 P

Fny= -1081.87 P P, then, Fex= +1081.87 P

Magnitude of the equilibrant (Fe)

$F_{n} = \sqrt{(F_{nx})^{2} +(F_{ny})^{2} }$

$F_{e} =\sqrt{(2341.87)^{2}+(1081.87)^{2} }$

Fe= 2579.68 P

Calculation of the direction of equilibrant force (α)

$\alpha =tan^{-1} (\frac{F_{ny} }{F_{nx} } )$

$\alpha =tan^{-1} (\frac{1081.87 }{2341.87} )$

α= 24.8°

Look at the attached graphic

6 0

3 years ago

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. the wire ful

il63 [147K]

Answer:

1.875*10^-5

Explanation:

8 0

3 years ago

Write the nuclear reaction equation for the beta decay of Iodine-131

den301095 [7]

Answer:

$_{53}^{131}I \rightarrow _{54}^{131}Xe + e + \bar{\nu}$

Explanation:

In a beta (minus) decay, a neutron in a nucleus turns into a proton, emitting a fast-moving electron (called beta particle) alongside with an antineutrino.

The general equation for a beta decay is:

$^A_Z X \rightarrow _{Z+1}^AY+^0_{-1}e+ ^0_0\bar{\nu}$ (1)

where

X is the original nucleus

Y is the daughter nucleus

e is the electron

$\bar{\nu}$ is the antineutrino

We observe that:

The mass number (A), which is the sum of protons and neutrons in the nucleus, remains the same in the decay
The atomic number (Z), which is the number of protons in the nucleus, increases by 1 unit

In this problem, the original nucles that we are considering is iodine-131, which is

$_{53}^{131}I$

where

Z = 53 (atomic number of iodine)

A = 131 (mass number)

Using the rule for the general equation (1), the dauther nucleus must have same mass number (131) and atomic number increased by 1 (54, which corresponds to Xenon, Xe), therefore the equation will be:

$_{53}^{131}I \rightarrow _{54}^{131}Xe + e + \bar{\nu}$

7 0

3 years ago

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