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3 years ago

Please I need your help

Physics

1 answer:

frez [133]3 years ago

5 0

1.14 km = Distance
2.30 m/s = Speed
5.12 cm/s2 = Speed
6.150 mph = Distance
8.3.2 sec = Speed
9.25 ft = Distance

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A cannon fires a cannonball 500.0 m downrange when set at a 45.0o angle. At what velocity does the cannonball leave the cannon?

kobusy [5.1K]

Answer:

v = 70 m/s

Explanation:

Range on the cannon ball is given as

$d = 500.0 m$

here the angle of the projection of the ball is given as 45 degree

now we know that if the velocity of the ball is "v" then its two components will be given as

$v_x = vcos45$

$v_y = vsin45$

so here time of flight of the motion is given as

$T = \frac{2v_y}{g}$

$T = \frac{2vsin45}{g}$

also the range is given as

$R = v_x T$

$R = (vcos45)(\frac{2vsin45}{g})$

now plug in all data in this equation

$500.0 = \frac{v^2(2sin45cos45)}{g}$

$v = 70 m/s$

4 0

3 years ago

Suppose a ray of light traveling in a material with an index of refraction na reaches an interface with a material having an ind

ivolga24 [154]

Answer: Option (b) is the correct answer.

Explanation:

It is known that when a ray of light tends to travel from a denser to rarer medium then there occurs total internal reflection.

For a denser medium the refractive index is greater than that of a rarer medium. This means that for the given situation refractive index of medium $n_{a}$ is greater than medium $n_{b}$ .

this also means that incident angle must be greater than the critical angle of the medium.

Thus, we can conclude that the statement $n_{a}$ > $n_{b}$ must be true for total internal reflection to occur.

5 0

3 years ago

Use the graph above to determine the change in speed of the object between 20 and 30 seconds?

allsm [11]

Answer:

6 m/s

Explanation:

To determine the change in speed of the object, we just need to determine its speed at t = 30 s and at t = 20 s, and then calculate the difference.

The speed at t = 30 is:

v = 6 m/s

While the speed at t = 20 s is:

u = 0

Therefore, the change in speed is:

$\Delta v = v-u=6-0 = 6 m/s$

4 0

3 years ago

Read 2 more answers

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The sprin

iogann1982 [59]

Answer:

Therefore, the spring constant of each spring = 1.6 × 10⁻⁶ kg/s².

Explanation:

The period (T) of a spring in oscillation = 2π √(m/k)............. equation 1

Where m = mass acting on the spring (kg), k = spring constant of the spring (kg/s²).

Making k the subject of equation 1

k = T²/(4π²×m) .......................... equation 2

From the question, F = 4.42 Hz,

since T = 1/F

then, T = 1/F = 1/4.42 =0.226 s, π = 3.143

since the weight of the mass is evenly distributed over the four identical spring, Hence

m = 1450/4 = 362.5 kg

Substituting these values into equation 2

k = 0.226/{(4×3.143²)362.5}

k = 0.226/(14323.751)

k = 0.0000016 kg/s²

k = 1.6 × 10⁻⁶ kg/s².

Therefore, the spring constant of each spring = 1.6 × 10⁻⁶ kg/s².

5 0

3 years ago

What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$