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BlackZzzverrR [31]
3 years ago
7

Assume the ground is uniformly level. If the horizontal component a projectile's velocity is doubled, but the vertical component

is unchanged, what is the effect on the time of flight
Physics
1 answer:
Katarina [22]3 years ago
3 0

Answer:

Explanation:

Time of flight = 2 x u sinα / g where u sinα is vertical component of projectile's velocity u .

So Time of flight = 2 x vertical component / g

vertical component = constant

g is also constant so

Time of flight will also be constant .

It will remain unchanged .

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As a defensive measure, forces afloat would typically operate in highly dispersed formations, sometimes covering an area of more
hoa [83]

Answer:

B - False

Explanation: a dispersed formation means spreading out, more like an attacking or advancing position. This doesn't depict a defensive measure.

7 0
4 years ago
Two balloons having the same charge of 1.2 × 10-6 coulombs each are kept 5.0 × 10-1 meters apart. What is the magnitude of the r
ra1l [238]
Steps: 1)Data: q1=charge of first balloon=1.2 × 10^-6 q2=charge of second balloon=1.2 × 10^-6 Condition:q1=q2 then for convenience,we say q instead of q1 and q2. Radius=r=5*10^-1 Radius is the distance between balloons! Magnitude of repulsive force=F=? 2)Solution: F=kq1q2/r^2 As q1=q2=q So F=kqq/r^2 =kq^2/r^2 Putting values: F=((9*10^9)(1.2 × 10^-6)^2)/(5*10^-1)^2 Using calculator: F=0.05184 N
4 0
3 years ago
Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision
Sidana [21]

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

Learn more about conservation of momentum principle

brainly.com/question/14033058

#SPJ2

4 0
2 years ago
Read 2 more answers
Mr. White claims that he invented a heat engine with a maximum efficiency of 90%. He measured the temperature of the hot reservo
Studentka2010 [4]

Answer:

The error he made was that he didn't convert the unit of temperature to Kelvin.

The correct efficiency is 24%

Explanation:

Parameters given:

Temperature of hot reservoir = 100°C = 373 K

Temperature of cold reservoir = 10°C = 273 K

The efficiency of a heat engine is given as:

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Where

Qc = Output heat;

Qh = Input heat;

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Th = Temperature of the hot reservoir.

=> E = 1 - (283/373)

E = 1 - 0.76

E = 0.24

In percentage,

E = 0.24 * 100 = 24%

Hence, the efficiency of the engine is actually 24%.

The error he made was that he didn't convert the temperature to Kelvin. If we leave the temperatures in °C, we have that:

E = 1 - (10/100)

E = 1 - 0.1 = 0.9

In percentage,

E = 0.9 * 100 = 90%

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3 years ago
Gravity does not actually "pull" objects at
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I believe it’s (D. Any object)
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