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3 years ago

What is question number 7. ?

Physics

1 answer:

Sergio039 [100]3 years ago

4 0

Answer:

initial velocity=0

final velocity=73 ms-1

mass of golf ball=62 g

F=m[vf-vi]/t

F=[62x10^-3][73-0]/[2x10^-3]

F=2263 N

Explanation:

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Arrange the distances between Earth and various celestial objects in order from least to greatest. Use the conversion table to h

Kaylis [27]

distance to the star Betelgeuse: 640 ly

As we know that

$1 ly = 63000 AU$

also we know that

$1AU = 1.5 \times 10^8 km$

$1 ly = 63000 (1.5 \times 10^8) = 9.45 \times 10^{12} km$

So the distance of Betelgeuse = 640 ly

$d_1 = 640 \times 9.45 \times 10^{12} = 6.05 \times 10^{15} m$

distance to the star VY Canis Majoris: $3.09 × 10^8 AU$

$d_2 = 3.09\times 10^8 \times 1.5 \times 10^8 km$

$d_2 = 4.64 \times 10^{16} km$

distance to the galaxy Large Magellanic Cloud: 49976 pc

$1 pc = 3.262 ly = 3.262 \times 9.45 \times 10^{12} km$

$1pc = 3.08 \times 10^{13} km$

now we have

$d_3 = 49976 \times 3.08 \times 10^{13}$

$d_3 = 1.54 \times 10^{18} km$

distance to Neptune at the farthest: 4.7 billion km

$d_4 = 4.7 \times 10^9 km$

now the order of distance from least to greatest is as following

1. distance to Neptune at the farthest

2. distance of Betelgeuse

3. distance to the star VY Canis Majoris

4. distance to the galaxy Large Magellanic Cloud

6 0

3 years ago

A sprinter accelerates from rest to 10.0 m/s in 1.28 s . Part A Part complete What is her acceleration in m/s2? a a = 7.81 m/s2

Mashutka [201]

Explanation:

It is given that,

Initial speed of sprinter, u = 0

Final speed of sprinter, v = 10 m/s

Time taken, t = 1.28 s

a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{10\ m/s}{1.28\ s}$

$a=7.81\ m/s^2$

b. Final speed of the sprinter, v = 36 km/h

Time, t = 0.000355 h

Acceleration, $a=\dfrac{36}{0.000355}$

$a=101408.45\ km/h^2$

Hence, this is the required solution.

3 0

4 years ago

A 2 kg ball is dropped above the surface of Planet X. If the gravitational field strength at the surface of Planet X is 5 N/kg,

Trava [24]

Given data:

* The mass of the ball is 2 kg.

* The gravitational field strength at the surface of planet X is 5 N/kg.

Solution:

The weight of the ball on the planet X is,

$W=ma$

where m is the mass of ball, a is the gravitational field strength,

Substituting the known values,

$\begin{gathered} W=2\times5 \\ W=10\text{ N} \end{gathered}$

Thus, the weight of the ball on the surface of planet X is 10 N.

3 0

1 year ago

Kissa and Lily ran a race in gym class Lily was off balance so she did not start until 2 seconds had passed

Paul [167]

Average speed = (distance covered) / (time to cover the distance)

Tissa covered 60 meters in 10 seconds. Her average speed was

(60 m) / (10 sec) = 6 m/s.

That's the slope of the dotted line.

Lilly covered 60 meters in 8 seconds. Her average speed was

(60 m) / (8 sec) = 7.5 m/s .

That's the slope of the solid line.

Lilly covered the same distance in less time, and both girls
arrived at the finish line together. Technically, in science talk,
we would say that Lilly ran "faster", and her average speed
was "greater".

We can detect that by looking at the graph, because Lilly's line
has the characteristic of being "steeper", and we know that the
slope of the line on a distance/time graph is "speed".

3 0

3 years ago

A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three ne

Masteriza [31]

Answer:

The electric flux is zero because charge is zero.

Explanation:

Given that,

Positive charge $q_{1}=3.90\times10^{-12}\ C$