Which is NOT a way to stay safe from static electricity?

A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sendin

Lelechka [254]

Answer:

Explanation:

Given

mass of boy $m_b=70.9\ kg$

mass of girl $m_g=43.2\ kg$

speed of girl after push $v_g=4.64\ m/s$

Suppose speed of boy after push is $v_b$

initially momentum of system is zero so final momentum is also zero because momentum is conserved

$P_i=P=f$

$0=m_b\cdot v_b+m_g\cdot v_g$

$v_b=-\frac{m_g}{m_b}\times v_g$

$v_b=-\frac{43.2}{70.9}\times 4.64$

$v_b=-2.82\ m/s$

i.e. velocity of boy is 2.82 m/s towards west

8 0

3 years ago

The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of

kipiarov [429]

Answer:

Δh_air=714m

Explanation:

Given data

$P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\ g=9.8m/s^{2}$

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

=(Δh_air)× p_air ×g

Then

Δh_air=(pHg+ΔhHg)÷p_air

$=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m$

Δh_air=714m

7 0

3 years ago

Yashoda prepares some lime juice on a hot day. She adds 80 g of ice at a temperature of 0°C to 0.32 kg of lime juice. The temper

Vikentia [17]

Answer:

Explanation:

a )

hear energy required to melt 1 g of ice = 340 J ,

hear energy required to melt 80 g of ice = 340 x 80 J = 27220 J .

b ) energy gained by the melted ice ( water at O°C ) = m ct

where m is mass of water , s is specific heat and t is rise in temperature

= 80 x 4.2 x ( 8°C - 0°C)

= 2688 J .

c )

energy lost by lime juice = energy gained by ice and water

= 27220 J + 2688 J .

= 29908 J .

d )

Let specific heat required be S

Heat lost by lime juice = M S T

M is mass of lime juice , S is specific heat , T is decrease in temperature

= 320 g x S x ( 29 - 8 )°C

= 6720 S

For equilibrium

Heat lost = heat gained

6720 S = 29908 J

S = 4.45 J /g °C .

4 0

3 years ago

Science: Which statement is true about star A

Elza [17]

Answer:

no clue

Explanation:

6 0

3 years ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

IRINA_888 [86]

Answer:

Time : 7.96 s

Distance Traveled : 357.8 m

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

t₂ = 7.96 s

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

S = 357.8 m

7 0

3 years ago