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prohojiy [21]
3 years ago
14

Help me please I need to turn it in before midnight...

Physics
1 answer:
ale4655 [162]3 years ago
7 0

Answer:

The statement which explains how the total time spent in the air is affected as the projectile's angle of launch increases from 25 degrees to 50 degrees is;

C. Increasing the angle from 25° to 50° will increase the total time spent in the air

Explanation:

The equation that can be used to find the total time, T, spent in the air of a projectile is given as follows;

T = \dfrac{2 \cdot u \cdot sin(\theta) }{g}

Where;

T = The time of flight of the projectile = The time spent in the air

u = The initial velocity of the projectile

θ = The angle of launch of the projectile

g = The acceleration due to gravity ≈ 9.81 m/s²

Given that sin(50°) > sin(25°), when the angle of launch, θ, is increased from 25 degrees to 50 degrees, we have;

Let T₁ represent the time spent in the air when the angle of launch is 25°, and let T₂ represent the time spent in the air when the angle of launch is 50°, we have;

T_1 = \dfrac{2 \cdot u \cdot sin(25^{\circ}) }{g}

T_2 = \dfrac{2 \cdot u \cdot sin(50^{\circ}) }{g}

sin(50°) > sin(25°), therefore, we have;

\dfrac{2 \cdot u \cdot sin(50^{\circ}) }{g} >   \dfrac{2 \cdot u \cdot sin(25^{\circ}) }{g}

Therefore;

T₂ > T₁

Therefore, increasing the angle at which the projectile is launched from 25° to 50° will increase the total time spent in the air.

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A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f
konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m

(c) the maximum speed attained by the object during its motion

E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s

8 0
3 years ago
Kamal said the distance from the top of the balloon to the ground in the Example is Ï···· 353 ft. What mistake might Kamal have
snow_lady [41]

Complete question is;

Kamal said the distance from the top of the balloon to the ground in the Example image attached is √353 ft. What mistake might Kamal have made?

Answer:

the mistake Kamal made is that she probably used 17 ft as the perpendicular side of the triangle with b as the hypotenuse instead of using 17ft as the hypotenuse

Explanation:

From the image attached, we can see that the distance from the top of the balloon which is blue in color to the ground is denoted by "b".

Now the triangle is a right angle triangle with hypotenuse = 15ft + 2ft = 17 ft; the adjacent side = 8 ft, while the opposite side is "b".

Thus, we can use pythagoras theorem to solve this as;

b = √(17² - 8²)

b = √(289 - 64)

b = √225

b = 15ft

However,we are told Kamal got b as √353 ft.

From inspection of the calculations we just did, if we had used addition instead of subtraction, we would have gotten b = √353 ft.

Thus, we can under that the mistake Kamal made is that she probably used 17 ft as the perpendicular side of the triangle with b as the hypotenuse instead of using 17ft as the hypotenuse.

7 0
3 years ago
Please I need help........
LenKa [72]

Answer:

7.46 J/kg/K

Explanation:

The heat absorbed or lost is:

q = mCΔT

where m is the mass, C is the heat capacity, and ΔT is the change in temperature.

Given q = 15.0 J, m = 0.201 kg, and ΔT = 10.0 °C:

15.0 J = (0.201 kg) C (10.0 °C)

C = 7.46 J/kg/°C

Which is the same as 7.46 J/kg/K.

7 0
3 years ago
Read 2 more answers
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
4 years ago
"
Volgvan

Answer:

The height is : 60.025 m

Explanation:

The flowerpot falls off the balcony with zero launch angle

Given the time of fright as 3.5 s then ;

The formula to apply is ;

T=\sqrt{\frac{2H}{g} }\\\\3.5=\sqrt{\frac{2H}{9.8} }

3.5²= 2H/9.8

12.25 =2H/9.8

12.25 * 9.8 = 2H

120.05 = 2H

120.05/2 = H

60.025 =H

7 0
3 years ago
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