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galben [10]
1 year ago
6

Isabella drops a pen off her balcony by accident while celebrating the successful completion of a physics problem. assuming air

resistance is negligible, how many seconds does it take the pen to reach a speed of 19.62 \,\dfrac{\text {m}}{\text s}19.62 âs â âm ââ 19, point, 62, space, start fraction, m, divided by, s, end fraction
Physics
1 answer:
expeople1 [14]1 year ago
8 0
U = 0, initial vertical velocity

Neglect air resistance, and g = 9.8 m/s².

The time, t, required for the pen to attain a vertical velocity of 19.62 m/s is given by
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s

Answer:  2.0 s
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The image shows positions of the earth and the moon in which region would an astronaut feel the lightest
trapecia [35]

Answer:

The moon region

Explanation:

This is because there is little to no gravity on the moon. That is where the astronaut would feel the lightest.

5 0
1 year ago
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A 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Afterwards, the 355 kg car moves right at 5.80
Sidana [21]

Explanation:

It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

202\ kg\times 8.5\ m/s+355\ kg\times 0=m_1v_1+355\ kg\times 5.8\ m/s

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

7 0
2 years ago
a rod of some material 0.20 m long elongates 0.20 mm on heating from 21 to 120°c. determine the value of the linear coefficient
Rufina [12.5K]

Answer:

The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹

Explanation:

Li = 0.2m

ΔL = 0.2 mm = 0.0002m

T1 = 21ºC

T2 = 120ºC

ΔT =99ºC

α =ΔL/(Li*ΔT)

α =0.0002m /(0.2m * 99ºC)

α = 1.01 *10⁻⁵   (ºC)⁻¹

4 0
1 year ago
A typical raindrop is much more massive than a mosquito and falling much faster than a mosquito flies. How does a mosquito survi
seropon [69]

Answer:

Part a)

v = 7.94 m/s

Part b)

a = 992.6 m/s^2

Explanation:

Part a)

As we know that we can use momentum conservation for this

so we will have

m_1v_1 = (m_1 + m_2)v

(50m)8.1 = (50m + m)v

v = 7.94 m/s

Part b)

As we know that acceleration is rate of change in velocity

so we have

a = \frac{v_f - v_i}{t}

so we have

a = \frac{7.94 - 0}{8 \times 10^{-3}}

a = 992.6 m/s^2

5 0
1 year ago
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An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the sign
saul85 [17]

Answer:

2.1\times 10^{-9} T

Explanation:

We are given that

Frequency,f=800KHz=800\times 10^{3} Hz

1kHz=10^{3} Hz

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Electric field,E=0.63V/m

We have to find the magnetic field amplitude of the signal at that point.

c=3\times 10^8 m/s

We know that

B=\frac{E}{c}

B=\frac{0.63}{3\times 10^8}=0.21\times 10^{-8} T

B=2.1\times 10^{-9} T

8 0
2 years ago
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