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3 years ago

Which of the following has the most momentum?

Physics

1 answer:

777dan777 [17]3 years ago

5 0

Answer:

0.50 kg moving at 10 m/s

Explanation:

mv = p

0.50(10) = 5.0 kg•m/s

5.0(0.50) = 2.5 kg•m/s

1(2) = 2 kg•m/s

0.40(10) = 4.0 kg•m/s

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Why might the 1990 earthquake and volcanic eruption in Luzon be related?

Art [367]

I am not sure but i think the answer is C

6 0

3 years ago

A 52 kg child on a swing is travelling at 6 m/s . What is his gravitational potential energy if he has 1000 J of the mechanical

DiKsa [7]

Answer:

The correct answer is "64 J".

Explanation:

The given values are:

Mass,

m = 52 kg

Velocity,

v = 6 m/s

Mechanical energy,

= 1000 J

Now,

The gravitational potential energy will be:

⇒ $P.E=1000-\frac{1}{2}mv^2$

$=1000-\frac{1}{2}\times 52\times (6)^2$

$=1000-26\times 36$

$=1000-936$

$=64 \ J$

7 0

3 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

3 years ago

20.0 m [N] - 15 m [S20degreesE]

denis-greek [22]

Answer:

thank for making me give up on life

Explanation:

I thought the stuff I had was hard wth is even that

3 0

3 years ago

A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa

pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

M= I A

M= $16 \times 10^{-3} \times \pi \times r^2$

M= $16 \times 10^{-3} \times \pi \times 0.3024^2$

M= $4.59 \times 10^{-3}\ A m^2$

b) torque exerted in the loop

$\tau = M\ B$

$\tau = 4.59 \times 10^{-3}\times 0.79$

$\tau = 3.63 \times 10^{-3} N.m$

8 0

3 years ago