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Natali [406]
3 years ago
7

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. What is the time necessary for the projec

tile to reach the ground below?
Physics
1 answer:
baherus [9]3 years ago
5 0
The horizontal speed has no effect on the answer. 

It doesn't matter whether you flick a marble horizontally from the roof,
fire a high-power rifle horizontally from the roof, drive a school bus straight
off the roof, or drop a bowling ball from the roof with zero horizontal speed. 
Their vertical speed is completely determined by gravity, (and it happens to
be the same for all of them).

Handy dandy formula for the distance covered by anything that starts out
with zero speed and accelerates to the end:

            Distance = (1/2) (acceleration) x (time)²

If the beginning of the journey is on Earth, then the acceleration is
9.8 m/s² ... the acceleration of gravity on Earth.  We'll assume that
the 55-meter rooftop in the question is part of a building on Earth.

                       55 meters  =  (1/2) (9.8 m/s²) x (time)²           

Divide each side
by  4.9 m/s² :            55 m / 4.9 m/s²  =  (time)²

                                 (time)²  =  (55/4.9)  sec²

Square-root
each side:                time  =  √(55/4.9 sec²)

                                           =      3.35 sec  .
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What are various systems of unit?​
Naya [18.7K]

Answer:

hlw its jess bregoli

your answer is here

SI (International System of Units) (meter-kilogram-second-ampere-kelvin-mole-candela)

FPS (foot-pound-second)

MKS (meter-kilogram-second)

CGS (centimeter-gram-second)

EMU (Electromagnetic) (centimeter-gram-second-abampere)

ESU (Electrostatic) (centimeter-gram-second-abcoulomb)

Atomic (bohr-electron mass-atomic second-electron)

MTS (meter-tonne-second)

Explanation:

hope it may help you !!

6 0
2 years ago
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If the color red has a wavelength of 7.0 107 m. its frequency is
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I think it’s 5.0.1014
7 0
3 years ago
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A mass of 0.34 kg is fixed to the end of a 1.4 m long string that is fixed at the other end. Initially at rest, he mass is made
frozen [14]

At time t seconds, the mass has angular speed

\omega = \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t

and hence linear speed

v = (1.4\,\mathrm m) \omega = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t

After 8 s, its linear speed is

v = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) (8\,\mathrm s) = 37.072 \dfrac{\rm m}{\rm s} \approx 37 \dfrac{\rm m}{\rm s}

and has centripetal acceleration with magnitude

a = \dfrac{v^2}{1.4\,\rm m} \approx 981.667\dfrac{\rm m}{\mathrm s^2} \approx 980 \dfrac{\rm m}{\mathrm s^2}

To maintain this linear speed, by Newton's second law the required centripetal force should have magnitude

F = (0.34\,\mathrm{kg}) a \approx 333.767\,\mathrm N \approx \boxed{330 \,\mathrm N}

5 0
2 years ago
Calculate their densties in Si unit.<br>200mg,0.0004m​
Kryger [21]

Question: calculate their densties in Si unit.

200mg,0.0004m​³

Answer:

0.5 kg/m³

Explanation:

Applying,

D = m/V........................ Equation 1

Where D = density, m = mass, V = volume.

From the question,

Given: m = 200 mg = (200/1000000) kg = 2.0×10⁻⁴ kg, V = 0.0004 m³ = 4.0×10⁻⁴ m³

Substitute these values into equation 1

D = (2.0×10⁻⁴ kg)/(4.0×10⁻⁴)

D = 2/4

D = 0.5 kg/m³

Hence the density in S.I unit is 0.5 kg/m³

3 0
3 years ago
As a 3.0 kg bucket is being lowered into a 10 m deepwell, starting from top, the tension in the rope is 9.8 N. theacceleration o
777dan777 [17]

Answer:

A) 6.5 m/s²

Explanation:

Mass of the bucket, m = 3.0 kg

depth of the well, d = 10 m

tension on the rope, T = 9.8 N

The net downward force on the bucket is given as;

T = mg - ma

where;

a is downward acceleration of the bucket

9.8 = (3 x 9.8) - 3a

9.8 = 29.4 - 3a

3a = 29.4 - 9.8

3a = 19.6

a = 19.6 / 3

a = 6.53 m/s² downwards

Therefore, the acceleration of the bucket is 6.53 m/s² downwards

8 0
3 years ago
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