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2 years ago

14

Mass can be expressed in a Kilograms b Grams C Milligram d All of the above

2 answers:

IgorC [24]2 years ago

5 0

Answer:

d

Explanation:

cause mass is in everything

trapecia [35]2 years ago

3 0

Answer:

d) All of the above

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What is the appropriate term for the movement of the orange liquid in the lava lamp animation?

Taya2010 [7]

Heat energy is responsible for it.It is because the waxy substance present in the lava lamp get hot at its bottom and cold on its top and this process is mostly like convection

5 0

3 years ago

Two sound waves (wave X and wave Y) are moving through a medium at the same speed. If wave X has a greater frequency than wave Y

Naya [18.7K]

Answer:

Wave X has a shorter wavelength.

Explanation:

It is given that, two sound waves (wave X and wave Y) are moving through a medium at the same speed.

Speed of a wave is given by :

$v=\nu\times \lambda$

Where,

$\nu$ is frequency

$\lambda$ is wavelength

The relation between the frequency and the wavelength is inverse. Wave X has greater frequency than Y, then wave X has a shorter wavelength.

Hence, the correct option is (B) "has a shorter wavelength".

8 0

3 years ago

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 jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction

Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

$v=\frac{10}{2}=5m/s\;\cdots(i)$

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, $d=vt$

From equation (i)

$\Rightarrow d=5t\;\cdots(ii)$

As the jogger starts from origin, so, the distance, $d$ , also represents the position of the jogger at the time $t$ s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

$d=5\times 5=25 m$

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

$d=5\times 15=75 m$

So, the jogger is at a distance of 75 m from the origin.

8 0

2 years ago

Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the

Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan $m=12 kg$

Inclination $\theta =30^{\circ}$

Tension $T=29 N$

coefficient of Friction $\mu =0.18$

Resolving Forces Along x axis

$F_x=T\cos \theta -f_r$

where $f_r=friction\ Force$

$F_y=mg-N-T\sin \theta$

since there is no movement in Y direction therefore

$N=mg-T\sin \theta$

and $f_r=\mu N$

Thus $F_x=T\cos \theta -\mu N$

$F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))$

$F_x=25.114-18.558$

$F_x=6.556 N$

Work done by applied Force is equal to change to kinetic Energy

$F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2$

$6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2$

$v_f^2=\frac{6.556\times 2.7\times 2}{12}$

$v_f^2=2.95$

$v_f=1.717 m/s$

8 0

3 years ago