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natima [27]
3 years ago
12

The driving force behind the formation of the solar system and planets is called

Physics
1 answer:
laiz [17]3 years ago
6 0
I think it is high pressure / gravity and high temperatures
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A baseball is thrown at an angle of 40.0° above
monitta
Vector trigonometry can be used for this problem. Since the horizontal component is 12 meters per second, this is technically the hypotenuse (actual initial velocity) multiplied to cosine of 40 degrees. Therefore, to find the hypotenuse, we must divide 12 by cosine 40degrees. cos(40)= 0.766, and 12/0.766 = approximately 15.664, therefore our answer is (3) 15.7 m/s
7 0
3 years ago
Read 2 more answers
Will mark the brainliest
AysviL [449]

Answer:

The impulse transferred to the nail is 0.01 kg*m/s.

Explanation:

The impulse (J) transferred to the nail can be found using the following equation:

J = \Delta p = p_{f} - p_{i}

Where:                                                                

p_{f}: is the final momentum

p_{i}: is the initial momentum

The initial momentum is given by:

p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}}

Where 1 is for the hammer and 2 is for the nail.

Since the hammer is moving down (in the negative direction):

v_{1_{i}} = -10 m/s

And because the nail is not moving:

v_{2_{i}}= 0                      

p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:

p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

Since the hammer and the nail are moving in the negative direction:

v_{1_{f}} = v_{2_{f}} = -9.7 m/s

p_{f} = -9.7 m/s(7 \cdot 10^{-3} kg + 0.25 kg) = -2.49 kg*m/s

Finally, the impulse is:

J = p_{f} - p_{i} = - 2.49 kg*m/s + 2.50 kg*m/s = 0.01 kg*m/s

Therefore, the impulse transferred to the nail is 0.01 kg*m/s.

I hope it helps you!                                                                                                                                                                                                                                                                                                                                                                                                                    

7 0
3 years ago
Which statement will be true if you increase the frequency of a periodic wave?
-BARSIC- [3]
"The number of waves per second will increase" is the statement among the choices given in the question that <span>will be true if you increase the frequency of a periodic wave. The correct option among all the options that are given in the question is the first option or option "A". I hope that the answer has helped you.</span>
6 0
3 years ago
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You need to use your cell phone, which broadcasts an 830 MHz signal, but you’re in an alley between two massive, radio- wave-abs
Korvikt [17]

Answer:Angular width of the electromagnetic wave after it emerges from between the buildings is 2.761 degree

Explanation:

The wavelength of the electromagnetic wave is

\lambda=\frac{c}{f} \\\lambda=\frac{3\times 10^8}{830\times 10^6} \\\lambda=0.361 m

Consider two buildings as a single slit, for single slit diffraction dark fringes are located an angle of

\theta _{p}=sin^{-1}\frac{\lambda}{a}

where \lambda is frequency, a is space between buildings.

For the first dark fringes

\theta _{1}=sin^{-1}\frac{\lambda}{a}\\\theta _{1}=sin^{-1}\frac{0.3614}{15}\\\theta _{1}=1.38^{0}

Angular width of the electromagnetic wave after it emerges from between the buildings is

\Delta\theta =2\theta _{1}\\\Delta\theta =2\times 1.38\\\Delta\theta =2.761^0

6 0
3 years ago
A 1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s.
iren [92.7K]

Explanation:

Given that,

Mass of the ball, m = 1.2 kg

Initial speed of the ball, u = 10 m/s

Height of the floor from ground, h = 32 m

(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 32}

v = -25.04 m/s (negative as it rebounds)

The impulse acting on the ball is equal to the change in momentum. It can be calculated as :

J=m(v-u)

J=1.2\times (-25.04-10)

J = -42.048 kg-m/s

(b) Time of contact, t = 0.02 s

Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :

J=\dfrac{F}{t}

F=J\times t

F=42.048\times 0.02

F = 0.8409 N

Hence, this is the required solution.

5 0
3 years ago
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