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kakasveta [241]
3 years ago
10

A large dog with a mass of 30 kg chases a car at 2 m/s. What is the magnitude of its momentum?

Physics
1 answer:
AlexFokin [52]3 years ago
6 0

Answer: 60 kg-m/s

Explanation: I just answered that question and got it correct.

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20 Points available for physics help
Sonja [21]
The second one is correct not sure about the first one sorry
8 0
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A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1
baherus [9]

Answer:

The  value of the power is   P_c  =  38.55 \  W

Explanation:

From the question we are told that

   The  power  rating P_{1000} =P_b=  52 \  W

    The frequency is  f = 1000 \  Hz

    The  frequency at which the sound intensity decreases  f_k  =  20 \  Hz

     The decrease in intensity is by \beta  =  1.3 dB

Generally the  initial intensity of the speaker  is mathematically represented as

     \beta_1 =  10 log_{10} [\frac{P_b}{P_a} ]

Generally the intensity of the speaker after it has been decreased is

       \beta_2 =  10 log_{10} [\frac{P_c}{P_a} ]

So

\beta_1-\beta_2 =  10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]

=>  \beta =  10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3

=>  \beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3

=>  \beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3

=> 10log_{10} [\frac{P_b}{P_c} ] = 1.3

=> log_{10} [\frac{P_b}{P_c} ] = 0.13

taking atilog of both sides

[\frac{P_b}{P_c} ] = 10^{0.13}      

=>[\frac{52}{P_c} ] = 10^{0.13}      

=>  P_c  =  \frac{52}{1.34896}

=>   P_c  =  38.55 \  W

   

3 0
3 years ago
A wave is incident on the surface of a mirror at an angle of 41° with the normal. what can you say about its angle of reflection
Umnica [9.8K]

It's angle of reflection must be 41 degrees

we know, by the first law of reflection that angle of incidence is always equal to angle of reflection..........

7 0
3 years ago
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In 1998, scientists discovered that the expansion of the universe has been accelerating.
Lynna [10]
Dang that’s a cool fact bro
8 0
3 years ago
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A ball of mass 1.6 kg is attached to the end of a massless string. A circus clown twirls the
erik [133]

Compute the ball's angular speed <em>v</em> :

<em>v</em> = (1 rev) / (2.3 s) • (2<em>π</em> • 180 cm/rev) • (1/100 m/cm) ≈ 4.917 m/s

Use this to find the magnitude of the radial acceleration <em>a</em> :

<em>a</em> = <em>v </em>²/<em>R</em>

where <em>R</em> is the radius of the circular path. We get

<em>a</em> = <em>v</em> ² / (180 cm) = <em>v</em> ² / (1.8 m) ≈ 13.43 m/s²

The only force acting on the ball in the plane parallel to the circular path is the tension force. By Newton's second law, the net force acting on the ball has magnitude

∑ <em>F</em> = <em>m</em> <em>a</em>

where <em>m</em> is the mass of the ball. So, if <em>t</em> denotes the magnitude of the tension force, then

<em>t</em> = (1.6 kg) (13.43 m/s²) ≈ 21 N

7 0
3 years ago
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