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4 years ago

What is the mass of a dog that weighs 382 N?(unit=kg)

Physics

1 answer:

tino4ka555 [31]4 years ago

4 0

Answer:

38.77 kg

Explanation:

The weight of an object is given by:

Weight = mg, g = acceleration due to gravity ( 9.8 m/s square)

and m is the mass.

So, in this case:

Weight of dog is 382 N, so, by putting the values of weight and g into the equation we will get the mass of dog.

382 = mass multiply with 9.8

so, the mass of the dog will be 38.77 kg

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A 480 g falcon reaches a speed of 75 m/s in a vertical dive. if we assume that the falcon speeds up under the influence of gravi

poizon [28]

Answer:

286.7 m

Explanation:

So we are assuming the PE of the falcon is converted to KE

KE = PE

1/2 (.480)(75)^2 = .480 (9.81)(h ) solve for h = 286.7 m

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2 years ago

If your speed slows down you will have what acceleration?

vitfil [10]

When speed slows down you have kinetic energy.

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4 years ago

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A proton moves with a velocity of v with arrow = (3î − 5ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

Allisa [31]

Answer:

The magnitude of the magnetic force this particle experiences is $2.6\times10^{-9}\ N$ .

Explanation:

Given that,

Velocity v= (3i-5j+k) m/s

Magnetic field B=(i+2j-k) T

We need to calculate the value $\vec{v}\times\vec{B}$

$(\vec{v}\times\vec{B})=3i+4j+11k$

We need to calculate the magnitude of the magnetic force this particle experiences

Using formula of magnetic force

$\vec{F}=q(\vec{v}\times\vec{B})$

Put the value into the formula

$\vec{F}=1.6\times10^{-19}\times(3i+4j+11k)$

$\vec{F}=(4.8i+6.4j+1.76k)\times10^{-19}$

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Hence, The magnitude of the magnetic force this particle experiences is $2.6\times10^{-9}\ N$ .

7 0

3 years ago

A 1.0 cm3 air bubble is released from the sandy bottom of a warm, shallow sea, where the gauge pressure is 1.6 atm. The bubble r

Zigmanuir [339]

Answer:

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3 years ago

A wye-connected load has a voltage of 480v applied to it. What is the voltage drop across each phase

rodikova [14]

Answer:

$Y_A=277.128 \angle 30v$

$Y_B=277.128 \angle (-150)v$

$Y_C=277.128 \angle (90)v$

Explanation:

From the question we are told that

Voltage $V_L_L =480v$

Generally in a case of Y_connection $V_p_ h$ is mathematical represented as

$V_p_h=\frac{V_l_l}{\sqrt{3}} \angle (\phi-30)v$

Generally voltage drop across phase A

$Y_A=\frac{408}{\sqrt{3}} \angle -(0-30)$

$Y_A=277.128 \angle 30v$

Generally voltage drop across phase B

$Y_B=277.128 \angle (-30-120)$

$Y_B=277.128 \angle (-150)v$

Generally voltage drop across phase C

$Y_C=277.128 \angle (-30+120)$

$Y_C=277.128 \angle (90)v$

3 0

3 years ago