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3 years ago

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Rosa uses the formula (vi cos)t to do a calculation. Which value is Rosa most likely trying to find? the vertical acceleration o

f a projectile launched at an angle the vertical acceleration of a projectile launched horizontally the horizontal displacement of a projectile launched at an angle the horizontal displacement of a projectile launched horizontally

2 answers:

Morgarella [4.7K]3 years ago

8 0

Answer:

The answer is C my dude :)

blagie [28]3 years ago

4 0

The horizontal displacement of a projectile launched at an angle

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An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what facto

Lemur [1.5K]

Answer:

The factor of the diameter is 0.95.

Explanation:

Given that,

Power of old light bulb = 54.3 W

Power = 60 W

We know that,

The resistance is inversely proportional to the diameter.

$R\propto\dfrac{1}{D}$

The power is inversely proportional to the resistance.

$P\propto\dfrac{1}{R}$

$P\propto D^2$

We need to calculate the factor of the diameter of the filament reduced

Using relation of power and diameter

$\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}$

Put the value into the formula

$\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}$

$\dfrac{D_{i}}{D_{f}}=0.95$

$D_{i}=0.95 D_{f}$

Hence, The factor of the diameter is 0.95.

7 0

3 years ago

Read 2 more answers

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

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4 years ago

1. What is true about all forces?a. They are unbalanced b. They involve more than one object c. They cause objects to moved. d.

Pachacha [2.7K]

the answer is most likely D

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3 years ago

The displacement vector A and B when added together , give the resultant vector R so that R= A+B use the data in the drawing and

Salsk061 [2.6K]

The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.

Let:

b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°

We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.

Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.

In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.

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3 years ago

Assume that a gravitational anomaly in the solar system has shifted a field of asteroids into Earth’s orbit, and the field is no

Mandarinka [93]

Answer:

An asteroid is a minor planet of the inner Solar System. Historically, these terms have been applied to any astronomical object orbiting the Sun.

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2 years ago