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3 years ago

14

in a football game, the kicker kicks a football a horizontal distance of 43 yards if the ball lands 3.9 seconds later, what is t

he balls horizontal velocity

1 answer:

erica [24]3 years ago

7 0

Answer:

10s

Explanation:

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Orange light of wavelength 0.61 µ m in air enters a block of glass with εr = 1.44. What color would it appear to a sensor embedd

77julia77 [94]

Answer:

$0.5083\ \mu m$

Explanation:

$\lambda_0$ = Actual wavelength = $0.61\ \mu m$

$\varepsilon_r$ = Relative permittivity = 1.44

The observed wavelength in the glass is given by

$\lambda=\dfrac{\lambda_0}{\sqrt{\varepsilon_r}}\\\Rightarrow \lambda=\dfrac{0.61}{\sqrt{1.44}}\\\Rightarrow \lambda=0.5083\ \mu m$

The wavelength lies in the range of green light.

Hence, the observed color of light is $0.5083\ \mu m$

8 0

3 years ago

Two charged point particle are located at two vertices of an equilateral triangle and the electric field is zero at the third ve

Debora [2.8K]

Answer:

Option E

Explanation:

In the presence of two point charges at the two vertices of an equilateral triangle, the resultant electric field at the third vertex due to these charges can not be zero whether the charges are identical or not.

The reason being that only of the x or y component of the field can be cancelled out in either case still the total field can't be reduced to zero.

This can only be achieved if another charge is present.

4 0

3 years ago

A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and

Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

$W=mg$

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

$B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N$

(c) 1524 N

The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

$B=\rho_a V g$

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

4 0

3 years ago

in the primitive yo-yo apparatus (figure 1), you replace the solid cylinder with a hollow cylinder of mass m , outer radius r ,

kirza4 [7]

The magnitude of the downward acceleration of the hollow cylinder is 6m/s^2.

Z = I α

T.R =1/2 M ( $R^{2}$ + $(R/2)^{2}$ )α

T.R = 1/2M 5 $R^{2}$ /4 α

T = 5Ma/8

Mg - T = Ma

Mg - 5Ma/8 = Ma

Mg= 5Ma/8 + Ma = 13Ma / 8

acceleration = 8g/13 = 6 m/s^2

The rate at which an object's velocity with respect to time changes is called its acceleration. The direction of the net force imposed on an item determines its acceleration in relation to that force. According to Newton's Second Law, the magnitude of an object's acceleration is the result of two factors working together

The size of the net balance of all external forces acting on that item is directly proportional to the magnitude of this net resultant force; the magnitude of that object's mass, depending on the materials from which it is built, is inversely related to its mass.

Learn more about acceleration here:

brainly.com/question/2303856

#SPJ4

4 0

1 year ago

A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t

andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = $\frac{r}{2}$

f = $\frac{0.40}{2} = 0.20 m$

Now,

m = - $\frac{v}{u}$

- 2 = - $\frac{v}{u}$

$\frac{v}{u}$ = 2 (2)

Now, by lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

v = $\frac{uf}{u - f}$ (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = $\frac{uf}{u - f}$

2 = $\frac{f}{u - f}$

2(u - 0.20) = 0.20

u = 0.30 m

6 0

3 years ago