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2 years ago

An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50

Physics

1 answer:

Setler79 [48]2 years ago

6 0

Answer:

very hard others will answer it

Explanation:

hard

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A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s,

Art [367]

Answer:

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

Explanation:

given,

length of ladder = 10 ft

let x be the distance of the bottom and y be the distance of the top of ladder.

x² + y² = 100

differentiating with respect to time we get

$2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0$ ..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

$16\times 1.4 + 12\dfrac{dy}{dt} = 0$

$\dfrac{dy}{dt} = -\dfrac{5.6}{3}$

let the angle between the ladders be θ

$tan\theta = \dfrac{y}{x}$

y = xtan θ

$\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}$

$-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}$

$\dfrac{25}{2} \dfrac{d\theta}{dt} =\dfrac{-17.5}{6}$

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

6 0

2 years ago

Why is wood a renewable resource?

sweet-ann [11.9K]

Answer:

Trees grow everywhere, so we always have some

Explanation:

Wood is a renewable resource; trees can be replanted and grown to maturity in place of those that are cut down. As long as the trees are replanted at the same rate as they are cut down wood will be a renewable resource.

6 0

2 years ago

In the following reaction, which substance is the precipitate?(NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq

Delvig [45]

The precipitate is the barium sulfate, or 3) BaSO4. This is because it is a solid (as seen by the (s)), unlike the other aqueous product. The fact that it is a solid means that it is insoluble in water and therefore a precipitate.

Hope this helps!

7 0

2 years ago

Read 2 more answers

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

Anarel [89]

Answer:

$R = 24.3 m$

Explanation:

As we know that the orbital speed is given as

$v = \sqrt{\frac{GM}{R + h}}$

here we know that

v = 5500 m/s

$R = 4.48 \times 10^6 m$

$h = 630 km$

now we have

$5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}$

$M = 2.34 \times 10^24 kg$

now acceleration due to gravity of planet is given as

$a = \frac{GM}{R^2}$

$a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}$

$a = 7.7 m/s^2$

now range of the projectile on the surface of planet is given as

$R = \frac{v^2 sin2\theta}{g}$

$R = \frac{14.6^2 sin(2\times 30.8)}{7.7}$

$R = 24.3 m$

3 0

3 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

2 years ago