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4 years ago

You feel a static electric shock when free electrons build up on your body then you touch a

Physics

1 answer:

Arlecino [84]4 years ago

6 0

Unit: 10 TEST
4. A- charger

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a 7 kilogram cat is resting on top of a bookshelf that is 2 meters high. what is the cat’s gravitational potential energy relati

attashe74 [19]

PE = mgh
Mass, m = 7kg, g ≈ 10 m/s², height = 2m

= 7*10*2

= 140 Joules.

5 0

3 years ago

Read 2 more answers

What are electric motors?

Lapatulllka [165]

It's an electric motor that converts electric energy into Mechanical energy

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3 years ago

19. A 0.0340 kg bullet traveling at 120 m/s embeds itself in a 1.24 kg wooden block which

Blababa [14]

Answer: x ≈ 36.3 cm

Explanation:

Conservation of momentum during the collision

0.0340(120) + 1.24(0) = (0.0340 + 1.24) v

v = 3.2025 m/s

The kinetic energy of the block/bullet mass will convert to spring potential

½kx² = ½mv²

x = √(mv²/k)

x = √(1.274(3.2025²) / 99.0)

x = 0.363293... ≈ 36.3 cm

3 0

3 years ago

2) The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. How long doe

ValentinkaMS [17]

Answer: 2.04 s

Explanation:

Let the initial velocity be v, Angle of projectile be

Then the horizontal component = v cos θ = 16 m/s

Vertical component of velocity = v sin θ = 20 m/s

Time taken to reach the highest point is half the time taken for total flight.

Time for total flight,

$t = \frac{2vsin \theta}{g}$

$t'=\frac{vsin \theta}{g} = \frac {20 m/s}{9.8 m/s^2} = 2.04 s$

Thus, the football takes 2.04 s to rise to the highest point of its trajectory.

5 0

3 years ago

Use the definition of scalar product, a overscript right-arrow endscripts times b overscript right-arrow endscripts = ab cos θ,

makkiz [27]

Answer: $\theta=cos^{-1}0.991=7.69^o$

The following vectors have been given: $\vec{a}=3.0\widehat{i}+3.0\widehat{j}+3.0\widehat{k}\\ \vec{b}=5.0\widehat{i}+7.0\widehat{j}+6.0\widehat{k}$

The angle between these two vectors can be found by:

$cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\
||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}$

$\vec{a}.\vec{b}=a_xb_x+a_yb_y+a_zb_z\\ \vec{a}.\vec{b}=3\times5+3\times7+3\times6=15+21+18=54$

$||\vec{a}||=\sqrt{3^2+3^2+3^2}=\sqrt{27}\\ ||\vec{b}||=\sqrt{5^2+7^2+6^2}=\sqrt{110}$

$cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o$

7 0

3 years ago