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2 years ago

57. Example of the law of force and acceleration

Physics

1 answer:

7nadin3 [17]2 years ago

6 0

Answer:

Newton's Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass (object). Riding your bicycle is a good example of this law of motion at work. Your bicycle is the mass. Your leg muscles pushing pushing on the pedals of your bicycle is the force.

Explanation:

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a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

5 0

3 years ago

You have a 78.7 mF capacitor initially charged to a potential difference of 11.5 V. You discharge the capacitor through a 3.03 Ω

Aloiza [94]

Answer:

$\tau=0.23\;second$

Explanation:

Given,

$C=78.7\;mF\\V=11.5\;V\\R=3.03\;\Omega\\$

Time constant

$\tau=RC\\\tau=78.7\times10^{-3}\times3.03\\\tau=238.461\times10^{-3}\;second\\\tau=0.23\;second$

8 0

3 years ago

3. If the distance of the screen is moved from 100. cm to 200. cm the area of light would in-crease from 150. cm^2 to

german

Answer:

3) C

4 D

5) C

Explanation:

3) given that

Initial distance of the screen = 100cm

Initial area = 150 cm^2

Final distance = 200 cm

The intensity of light is inversely proportional to the square of the distance. That is

Intensity of light I = 1/d2

And also I = P/A

1/d^2 = P/A

P = A/d^2

P1 = P2

150/100 = A/200

1.5 = A/200

A = 1.5 × 200

A = 300 cm^2

4.) Light is projected onto a screen 75.0 cm from a light source. The light intensity = 4436 lux

If the screen is moved from 75.0 cm to 150. cm, the light sensor reading will be

Using inverse square law

I = 1/d^2

I×d^2 = constant. Therefore,

4436 × 75^2 = I × 150^2

I = 24952500/22500

I = 1109 lux

5.) We can express the relationship between luminosity, brightness, and distance with a simple formula.

As we tilt the serene the area of light decreases and makes the light more concentrated.

5 0

3 years ago

The x axis of a trajectory represents its ____________.

lubasha [3.4K]

The x axis of the trajectory will show the displacement in X direction only

So it is the Horizontal range

Correct answer would be

<em>Displacement in the horizontal direction(my guess)</em>

The stone is projected horizontally at height 64.8 m

now we can use kinematics in Y direction

$\Delta y = v_y * t +\frac{1}{2} gt^2$

now we have

$64.8 = 0 + \frac{1}{2}*9.8*t^2$

$t = 3.64 s$

<em>so here it will fall in the air for time t = 3.64 s</em>

4 0

3 years ago

Which is the fastest moving component in the electropherogram?

bazaltina [42]

An electropherogram, or electrophoregram, is a record or chart produced when electrophoresis is used in an analytical technique, primarily in the fields of molecular biology or biochemistry.

...

These genotypes can be used for:

genealogical DNA testing.

DNA paternity testing.

DNA profiling.

phylogenetics.

population genetics

hope this helped :)

6 0

3 years ago

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