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3 years ago

A leaf is floating on the Surface of the water,what will happen to its movement? Explain

Physics

1 answer:

docker41 [41]3 years ago

4 0

Answer:

MRCORRECT has answered the question

Explanation:

surface tension of water helps creatures(mostly of insecta class such as water striders) to walk on water. . it also helps water to move up the xylem tissue ofhigher plants without breaking up

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One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious spa

Ludmilka [50]

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

7 0

3 years ago

The two-way table below gives the thousands of commuters in Massachusetts in 2015 by transportation method and one-way length of

forsale [732]

Answer:

wow uh what? hdfgdfgf

Explanation:?

7 0

2 years ago

Read 2 more answers

When the distance between two masses is doubled, the gravitational attraction between

mrs_skeptik [129]

The new gravitational attraction will be 1/4 as much

Explanation:

The magnitude of the gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, the original force between the two objects is F, when they are separated by a distance r.

Later, the distance between the two objects is doubled, so the new distance is

$r'=2r$

Therefore, the new force will be

$F'=G\frac{m_1 m_2}{(2r)^2}=\frac{1}{4}(\frac{Gm_1 m_2}{r^2})=\frac{F}{4}$

Therefore, the new force will be one-fourth as much.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

5 0

3 years ago

a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg

IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

$M = \frac{q}{p}$

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

$4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m$

Now using thin lens formula:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\$

6 0

3 years ago

A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ

valentinak56 [21]

Explanation:

The given data is as follows.

mass = 0.20 kg

displacement = 2.6 cm

Kinetic energy = 1.4 J

Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

Total energy = Kinetic energy + spring potential energy

= 1.4 J + 2.2 J

= 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So, K.E = Total energy

= 3.6 J

Also, we know that

K.E = $\frac{1}{2}mv^{2}_{m}$

or, v = $\sqrt{\frac{2K.E}{m}}$

= $\sqrt{2 \times 3.6 J}{0.2 kg}$

= $\sqrt{36}$

= 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0

3 years ago