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Elena-2011 [213]
3 years ago
5

A book is thrown downward from the library window with a speed of 2.0\,\dfrac{\text m}{\text s}2.0 s m ​ 2, point, 0, start frac

tion, start text, m, end text, divided by, start text, s, end text, end fraction and lands on the ground 5.0\, \text m5.0m5, point, 0, start text, m, end text below. We can ignore air resistance. What is the final velocity of the book in \dfrac{\text m}{\text s} s m ​ start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?
Physics
2 answers:
dem82 [27]3 years ago
7 0

Answer: final Velocity v = 10.2m/s

Explanation:

Final speed v(t) is given as

v(t) = u + at .......1

Where; u = the initial speed

a = acceleration

t = time taken

The total distance travelled d is given as

d = ut + 1/2(at^2)

Given

d = 5.0m

u = 2.0m

a = g = 10m/s2 (acceleration due to gravity)

Substituting into the equation above we have

5 = 2t + 5t^2

5t^2 +2t -5 = 0

Applying the quadratic formula. We have;

t = 0.82s & t = -1.22s

t cannot be negative

t = 0.82s

From equation 1 above

v = 2.0m/s + 10(0.82)m/s

v = 10.2m/s

nadya68 [22]3 years ago
6 0

Answer:

-10

Explanation:

khan academy

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At an accident scene on a level road, investigators measure a car's skid mark to be 98 m long. It was a rainy day and the coeffi
LiRa [457]

Answer:

a. V = 19.1m/s

b. The mass of the car does not matter

Explanation:

A.

KE = 1/2mv² = fd --------(1)

Fd = umgd ---------(2)

Therefore,

1/2mv² = umgd ---------(3)

M will cancel itself out from both sides of equation 3.

Then we will have:

1/2v² = ugd

Then we cross multiply to make v² the subject of the formula

V² = 2ugd

V = √2ugd -------(4)

U = 0.38

g = 9.81

d = 98

When we input these values into equation 4, we will have:

V = √2x0.38x9.81x98

V = √730.6488

V = 27.03m/s

B.

The mass of the car does not actually matter as the mass was cancelled out on the both sides of equation 3

6 0
3 years ago
Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential
kondor19780726 [428]

Answer:

Part A: The voltage across resistor R1 is approximately \rm 2.1 \; V.

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

<h3>Part A</h3>

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of R_1 + R_2 = 50 + 90 = 140\; \Omega. The voltage across the two resistor, combined, is equal to \rm 6\; V. Hence by Ohm's Law, the current through the circuit will be equal to \rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A.

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to \rm \dfrac{3}{70}\; A. Apply Ohm's Law (again) to find the voltage across R1:

V = I \cdot R = \dfrac{3}{70} \times 50 \approx \rm 2.1\; V.

<h3>Part B</h3>

Since the equivalent resistance is equal to R_1 + R_2, when the value of R_1 decreases, the equivalent resistance will also decrease. By Ohm's Law, I = \dfrac{V}{R}. When the value of the denominator ( decreases, the value of the quotient, I the current through the circuit, will increase.

<h3>Part C</h3>

Keep in mind that if two resistors are connected in series,

I(R_1) = I(\text{Circuit}) = I(R_2).

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

Again, since the two resistors are connected in series,

V(R_1) + V(R_2) = V(\text{Circuit}) = \rm 6 \; V,

when the voltage across R2 increases, the voltage across R1 will decrease.

4 0
3 years ago
How do scientists classify small objects in the solar system?
LiRa [457]
*size
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4 0
3 years ago
What is the electric current measured in
cluponka [151]

An ampere (AM-pir), or amp

6 0
2 years ago
A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio
Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  \frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon  \frac{\omega}{W_n})^2}]

where;

\epsilon = 0

W_n = \sqrt { \frac{k}{m}}

= \sqrt { \frac{100*32.2}{200}}

= 4.0124

replacing them into the above equation and making X the subject of the formula:

X = Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}

X = 0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}

X = 5.676*10^{-3} \ mm

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

8 0
3 years ago
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