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3 years ago

Shari uses the data in the table to predict how her weight on Venus would compare with her weight on Earth.

Physics

2 answers:

spin [16.1K]3 years ago

3 0

Answer : Her weight is lower on Venus because the acceleration due to gravity is lower.

Explanation :

Venus is also called as Earth's twin. This is because both the mass and the size of Earth and Venus are almost same. The acceleration due to gravity on earth is $9.8\ m/s^2$ while on Venus is $8.87\ m/s^2$ .

So, when Shari measure her weight on Venus she found her weight is lower on Venus. This is because the acceleration due to gravity is lower on the surface of Venus as compared to the Earth.

Since, $w=mg$

i.e. weight depends on g.

So, correct prediction is (b)

andrey2020 [161]3 years ago

3 0

Answer:

the answer is B

Her weight is lower on Venus because the acceleration due to gravity is lower

Explanation:

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If the impulse acting on an object is 100n waht is force acting on it if the time 1 s

lutik1710 [3]

force = 100 N

Impulse ( N.s ) : Force ( N ) * Time ( s )

⇒ 100 N.s = force * 1 s

⇒ force = 100/1

⇒ force = 100 N

8 0

3 years ago

An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far r from the center

Diano4ka-milaya [45]

Answer:

2023857702.507m

Explanation:

$f=\frac{GMm}{r^{2} }$

recall from newton's law of gravitation

G=gravitational constant

mshew=50g

melephant=5*10^3kg

rearth=radius of the earth 6400km or 6400000m

mearth= masss of the earth

Gm(shrew)m(earth)/r(earth)^2 = Gm(elephant)m(earth)/r^2

strike out the left hand side and right hand side variables

m(shrew)/r(earth)^2 = m(elephant)/r^2

r^2 = m(elephant).r(earth)^2 / m(shrew) .........make r^2 the subject of the equation

r^2= $(5*10^{3} *(6400000)^{2} )/.05$

r^2=40960000000000

r=2023857702.507m

4 0

3 years ago

Distance in scientific notation

n200080 [17]

Answer:

(See explanation for further information)

Explanation:

a) The distances of each planet with respect to the Sun is:

Mercury - $57.909 \times 10^{6}\,mi$

Venus - $67.240\times 10^{6}\,mi$

Earth - $92.960\times 10^{6}\,mi$

Mars - $141.600\times 10^{6}\,mi$

Jupiter - $483.800\times 10^{6}\,mi$

Saturn - $888.200\times 10^{6}\,mi$

Uranus - $1,787\times 10^{6}\,mi$

Neptune - $2,795\times 10^{6}\,mi$

b) The solutions are presented below:

A. The distance between Venus and Jupiter is:

$\Delta s = 483.800\times 10^{6}\,mi-67.240\times 10^{6}\,mi$

$\Delta s = 416.560\times 10^{6}\,mi$

B. The combined distance from the Sun is:

$\Delta s = 57.909\times 10^{6}\,mi + 67.240\times 10^{6}\,mi +92.960\times 10^{6}\,mi$

$\Delta s = 218.109\times 10^{6}\,mi$

Which is less than the distance from the Sun to Neptune ( $2,795\times 10^{6}\,mi$ ).

C. The new distance of Earth is $929.60 \times 10^{6}\,mi$ ( $929,600,000\,mi$ ). Saturn would be the closest planet to Earth, whose distance:

Scientific notation:

$\Delta s = 929.600\times 10^{6}\,mi-888.200\,\times 10^{6}\,mi$

$\Delta s = 41.4\times 10^{6}\,mi$

Standard notation:

$\Delta s = 929,600,000\,mi - 888,200,000\,mi$

$\Delta s = 41,400.000\,mi$

3 0

4 years ago

A ball is dropped from a height of 20 meters. At what helght does the ball have a velocity of 10 meters/second?

Sunny_sXe [5.5K]

Answer:

5.10 meters.

Explanation:

v²=u²+2gh

or, (10)²=(0)²+2×9.8×h

or, 19.6h=100

or, h=5.10 meters

Hope, this helps you.

8 0

3 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

The frequency changes by a factor of 0.27.

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = 0.27

7 0

3 years ago