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Zina [86]
3 years ago
6

Imagine a rock is dropped from the top of a tall building.After 2 seconds of falling the rock is instantaneous speed In approxim

ately 20m/s
Physics
1 answer:
Fantom [35]3 years ago
8 0

This is True.

Explanation:

initial velocity= Vi=0

final velocity= V

acceleration due to gravity= 9.8 m/s²

using the kinematic equation V= Vi+gt

V=0+(9.8)(2)

V= 19.6 m/s which is approximately equal to 20 m/s

Thus after 2 seconds, the instantaneous speed of the rock is approximately 20 m/s.

You might be interested in
A triangular plate with height 6 ft and a base of 7 ft is submerged vertically in water so that the top is 2 ft below the surfac
xenn [34]

Answer:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

\int\limits^a_bγhxdy

we know that γ=62.5 lb/ft^{3}

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}

when substituting the x and y-values given on the graph, we get that the slope is:

m=\frac{0-6}{7-0}=-\frac{6}{7}

once we got this slope, we can substitute it in the point-slope form of the equation:

y_{2}-y_{1}=m(x_{2}-x_{1})

which yields:

y-6=-\frac{6}{7}(x-0)

which simplifies to:

y-6=-\frac{6}{7}x

we can now solve this equation for x, so we get that:

x=-\frac{7}{6}y+7

with this last equation, we can substitute everything into our integral, so it will now look like this:

\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy

Now that it's all written in terms of y we can now simplify it, so we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}

By using the fundamental theorem of calculus we get:

62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]

When solving we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

6 0
3 years ago
A situation that restricts something
Ostrovityanka [42]
Answer:

Laws

Explanation:
i don’t have one but i hope ur doing well during this quarantine
7 0
3 years ago
A sound source emits 20.0 W of acoustical power spread equally in all directions. The threshold of hearing is 1.0 × 10-12 W/m2.
ahrayia [7]

Answer:

0.0018 W/m²

Explanation:

Power and intensity are related as:

I=\frac{P}{4\pi r^2}

P=  20.0 W (given)

r = 30.0 m (given)

I=\frac{20.0}{4\pi(30.0)^2}=0.0018 W/m^2

Intensity in decibels:

I(dB)=10log\frac{I}{I_o}\\=10log\frac{0.0018}{10^{-12}}=92.5 dB

7 0
3 years ago
1. What force is needed to accelerate the stone of
Leokris [45]

Answer:

i think its a

Explanation:

and next is 4

6 0
2 years ago
A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the
Paul [167]

Answer:

<em>a) 3344 N</em>

<em>b) 3344 N</em>

Explanation:

This is the complete question

1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected.  A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units.  B. What is the magnitude of the force of the truck on the car?

Mass of the car = 1100 kg

Mass of the truck = 2200 kg

Force exerted on the ground by the car = 5000 N

The total mass in the system = 1100 + 2200 = 3300 Kg

Total force in the system = 5000 N

Recall that the force in the system = mass x acceleration

therefore,

5000 = 3300 x a

Total acceleration in the system = 5000/3300 = 1.52 m/s^2

The force on the truck individually fro the car, will be the product of this acceleration and its mass

Force on the truck = 2200 x 1.52 = <em>3344 N</em>

b) Force on the car From the truck will be equal to this force but will act in the opposite direction.

Force on the car from the truck is <em>3344 N </em>

7 0
3 years ago
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