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katovenus [111]
4 years ago
13

- Across

Physics
2 answers:
postnew [5]4 years ago
6 0

Answer:

1# Citation

2# Supporting

3# copied, or plagiarised

4# Support it

5# opinion

6# Inference

brilliants [131]4 years ago
4 0

Answer:

3)quote

5) conclusion

6) infer

Explanation:

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You are helping two friends from our class with a physics problem where a cart is pushed up a ramp. In examining the motion of t
stiks02 [169]

Answer: Acceleration will have 2 components, vertical and horizontal.

Net-vertical component can be positive, zero or negative depending upon the magnitude of the upward component of the applied acceleration.

Net-horizontal acceleration will  be equal to the horizontal component of the applied acceleration.

Explanation:

Since acceleration is a vector quantity and the cart is being pushed up the ramp, the ramp would be at some angle to the horizontal and hence there will be vertical and horizontal components of acceleration.

<u>For vertical acceleration:</u>

If the magnitude of the upward component of the applied acceleration is greater than the value of the acceleration due to gravity then the net vertical acceleration will be upward because it will overtake the value of acceleration due to gravity.

In case the upward component of the applied acceleration is lesser than the value of the acceleration due to gravity then the net vertical acceleration will be downward.

<u>For horizontal acceleration:</u>

This component remains unaffected and is equal to the horizontal component of the applied acceleration because there is no other acceleration acting in the horizontal direction.

But the net acceleration will not be solely in the vertical or horizontal direction because the block has to move forward on the inclined ramp so there will always exist a horizontal and a vertical component making the net acceleration to parallel to the ramp in upward direction if the body is going up the ramp.

8 0
3 years ago
Read 2 more answers
This is a change in the position of a body with respect to time relative to a reference point.
dusya [7]

Answer: MOTION

Explanation:

motion is defined as the displacement of an object with respect to time relative to a stationary object (reference point). A good example of an object that can serve as a reference point includes: a tree or a building. The movement of a body at constant speed towards a particular direction at regular intervals of time can be determined and it's called uniform motion.

There are different types of motion, these includes: simple harmonic motion,

linear motion,

circular motion,

Brownian motion,

Rotatory motion

5 0
3 years ago
Read 2 more answers
A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?
Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length f=\frac{R}{2}=\frac{12}{2}=6cm

Now for mirror we know that \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

So \frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}

16.66-0.333=\frac{1}{v}

v = 0.750 m

Now magnification of the mirror is m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025

5 0
4 years ago
A 10,000-watt radio station transmits at 535 kHz. Determine the number of joules transmitted per second. 10,000 J/s 10 J/s 535 J
REY [17]
1 nanowatt  =  1 nanojoule/sec
1 watt  =  1 joule/sec
10 watts  =  10 joules/sec
100 watts  =  100 joules/sec
742.914 watts  =  742.914 joules/sec
1,000 watts  =  1,000 joules/sec
10,000 watts  =  10,000 joules/sec
100,000 watts  =  100,000 joules/sec
1 megawatt  =  1 megajoule/sec
1 gigawatt  =  1 gigajoule/sec
1 petawatt  =  1 petajoule/sec

We don't care what frequency the transmission is using,
or who their morning DJ is.

5 0
4 years ago
1) Si un mango cae a una velocidad de 75m/s y tarda 26 seg. en caer. ¿ Cuál habrá sido la velocidad con qué el mango llegó al su
Lyrx [107]

Answer:

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

Explanation:

El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:

v = v_{o}+g\cdot t (1)

Donde:

v_{o} - Velocidad inicial, en metros por segundo.

v - Velocidad final, en metros por segundo.

g - Aceleración gravitacional, en metros por segundo al cuadrado.

t - Tiempo, en segundos.

Si sabemos que v_{o} = -75\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} y t = 26\,s, entonces la velocidad final del mango es:

v = v_{o}+g\cdot t

v = -75\,\frac{m}{s}+\left(-9.807\,\frac{m}{s} \right)\cdot (26\,s)

v = -329.982\,\frac{m}{s}

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

8 0
3 years ago
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