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2 years ago

After a flood, people should guard against

Physics

2 answers:

aniked [119]2 years ago

7 0

It is c because there can be many diseases.

tia_tia [17]2 years ago

6 0

C because of the effect floods have

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After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 5.45 rad/s and it rotated 14.4 re

Komok [63]

Answer:

(a) α = -0.16 rad/s²

(b) t = 33.2 s

Explanation:

(a)

Applying 3rd equation of motion on the circular motion of the tire:

2αθ = ωf² - ωi²

where,

α = angular acceleration = ?

ωf = final angular velocity = 0 rad/s (tire finally stops)

ωi = initial angular velocity = 5.45 rad/s

θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad

Therefore,

2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²

α = -(29.7 rad²/s²)/(57.6π rad)

α = -0.16 rad/s²

Negative sign shows deceleration

(b)

Now, we apply 1st equation of motion:

ωf = ωi + αt

0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t

t = (5.45 rad/s)/(0.16 rad/s²)

t = 33.2 s

6 0

3 years ago

A car initially traveling at 27.2 m/s undergoes a constant negative acceleration of magnitude 1.90 m/s2 after its brakes are app

zubka84 [21]

Answer:

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

Explanation:

We can use the following equation:

$\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta$ (1)

The angular acceleration is:

$a_{tan}=\alpha R$

$\alpha=\frac{1.9}{0.325}$

$\alpha=5.85\: rad/s^{2}$

and the initial angular velocity is:

$\omega_{i}=\frac{v}{R}$

$\omega_{i}=\frac{27.2}{0.325}$

$\omega_{i}=83.69\: rad/s$

Now, using equation (1) we can find the revolutions of the tire.

$0=83.69^{2}-2*25.85 \Delta \theta$

$\Delta \theta=135.47\: rad$

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

I hope it helps you!

6 0

3 years ago

Problem 21-40a:

Greeley [361]

Answer:

Part a)

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

$q = 3.37 \times 10^{-4} C$

Explanation:

As we know that electric force on electric charge is given as

$F = qE$

here we have

$q = 1.6 \times 10^{-19}C$

E = 153 N/C

now force is given as

$F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N$

Gravitational force on electric charge near surface of earth is given as

$F_g = mg$

$F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N$

now the ratio of two forces is given as

$\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}$

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

$F_g = qE$

$mg = qE$

$(5.25 \times 10^{-3})(9.81) = q(153)$

here we have

$q = 3.37 \times 10^{-4} C$

8 0

3 years ago

A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th

solniwko [45]

Answer:

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

$a_t = R\alpha$

2) Centripetal acceleration

$a_c = \omega^2 R$

here we know that

$\omega = \alpha t$

$a_c = (\alpha t)^2 R$

now we know that net linear acceleration is given as

$a = \sqrt{a_c^2 + a_t^2}$

so we have

$a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}$

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

4 0

2 years ago

Elements in which family are most likely to have properties associated with

Elena-2011 [213]

Answer: As with all metals, the alkali metals are malleable, ductile, and are good conductors of heat and electricity. The alkali metals are softer than most other metals.

Alkaline earth metals

The alkaline earth elements are metallic elements found in the second group of the periodic table

Explanation:

5 0

3 years ago