Pre-Algebra and Physical Science

m_a_m_a [10]

Climatology is a subspecialty of Meterology.

3 0

2 years ago

The gravitational force strength on the moon is 1.63N/Kg if the rock on the moon weighs 200N how much does the same rock weigh o

FromTheMoon [43]

m = Q(on moon) * G(on moon) = 200N * 1.63N/kg = 326kg

Q(Earth)= g * m = 10m/s2 * 326kg = 3260N

8 0

2 years ago

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

2 years ago

The diagram below shows a circuit with a battery that provides an unknown potential difference. Work out the potential differenc

Gnesinka [82]

Answer:

no diagram attached :/

5 0

2 years ago

THE LENGTH OF A PENDULUM IS (1.5±0.01)m AND THE ACCELERATION DUE TO GRAVITY IS TAKEN AS (9.8±0.1)ms-² calculate the time period

tiny-mole [99]

Answer:

2.4583 ± 0.0207 seconds

Explanation:

The time period of a pendulum is approximately given by the formula ...

T = 2π√(L/g)

The maximum period will be achieved when length is longest and gravity is smallest:

Tmax = 2π√(1.51/9.7) ≈ 2.47903 . . . seconds

The minimum period will be achieved for the opposite conditions: minimum length and maximum gravity:

Tmin = 2π√(1.49/9.9) ≈ 2.43756 . . . seconds

If we want to express the uncertainty using a symmetrical range, we need to find half their sum and half their difference.

T = (2.47903 +2.43756)/2 ± (2.47903 -2.43756)/2

T ≈ 2.4583 ± 0.0207 . . . seconds

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We have about 2+ significant digits in the given parameters, so the time might be rounded to 2.46±0.02 seconds.

5 0

3 years ago