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kotegsom [21]
3 years ago
15

For a projectile launched at an angle, are the forces in the vertical direction balanced or unbalanced?

Physics
2 answers:
marissa [1.9K]3 years ago
8 0
Unbalanced because of the different forces pulling and pushing in certain directions
mart [117]3 years ago
7 0
Unbalanced as long as it is moving up or down. Immediately after being fired the vertical force from the launch will be greater than the force of gravity. for an instant at the the beginning of the projectiles decent the force will actually be balanced and there will be no vertical movement. After that the force of gravity is greater.

tl;dr they are unbalanced 
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3 years ago
A 0.89 m aqueous solution of an ionic compound with the formula mx has a freezing point of -3.0 ∘c . van't hoff factor?
BigorU [14]

Answer is: V<span>an't Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b -  molality, moles of solute per kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).

i = 1,81.

5 0
4 years ago
How is a coil of current carrying wire similar to a bar magnet
zalisa [80]

Answer:

 When an electric current flows, the shape of the magnetic field is very similar to the field of a bar magnet

Explanation:

5 0
3 years ago
an object of mass 8 kg is whriled round in a vertical circle of radius 2m with a constant speed of 6m/s .Then the maximum and mi
algol13

Answer:

Maximum Tension=224N

Minimum tension= 64N

Explanation:

Given

mass =8 kg

constant speed = 6m/s .

g=10m/s^2

Maximum Tension= [(mv^2/ r) + (mg)]

Minimum tension= [(mv^2/ r) - (mg)]

Then substitute the values,

Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N

Minimum tension= [8 × 6^2)/2 -(8×9.8)]

=64N

Hence, Minimum tension and maximum Tension are =64N and 2224N respectively

5 0
3 years ago
A mass M is suspended from a spring and oscillates with a period of 0.840 s. Each complete oscillation results in an amplitude r
iren [92.7K]

The energy becomes 0.50 times in 6.72 s.

Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.

Et=0.50E

1

k(4₂)² = (0.5) - kA²

(4₂)² = (0.5) A²

At = 0.71A

So, the amplitude of the oscillator becomes 0.71 times its initial ar

0.71A = = A(0.96)¹2

log(0.71)

log(0.96)

8.4

n=

So, the time taken for n oscillation is obtained as,

t = n (0.800 s)

= (8.4) (0.800)

= 6.72 s

learn more about oscillators brainly.com/question/15169199

#SPJ1

8 0
2 years ago
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