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3 years ago

Harvesting timber (wood) from a forest does not harm___________.

2 answers:

8 0

It does not harm the buildings or the cars in a nearby city. However, it does harm just about everything in the forest that's anywhere near the (former) tree.

Trava [24]3 years ago

6 0

Your answer would be soil. When you harvest timber (wood), you affect the wildlife around it, not the soil. Cutting down a tree won't make much of a difference to the soil underneath or around it.

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A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the poli

NNADVOKAT [17]

Answer:74.33 feet per sec

Explanation:

Given

Police car is 50 feet side of road

Red car is 140 feet up the road

Distance between them is decreasing at the rate of 70 feet per sec

From figure

$x^2+y^2=z^2$

$z=\sqrt{140^2+50^2}$

z=148.66 feet

as the red car is moving

therefore its velocity magnitude is given by

$\frac{\mathrm{d} x}{\mathrm{d} t}$

$2x\times \frac{\mathrm{d} x}{\mathrm{d} t}+0=2z\frac{\mathrm{d} z}{\mathrm{d} t}$

$x\frac{\mathrm{d} x}{\mathrm{d} t}=z\frac{\mathrm{d} z}{\mathrm{d} t}$

$140\times \frac{\mathrm{d} x}{\mathrm{d} t}=148.66\times 70$

$\frac{\mathrm{d} x}{\mathrm{d} t}=74.33 feet\ per\ sec$

6 0

3 years ago

a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance

aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b) Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

= $\frac{KQ}{r^{2} }$ for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{3.10^{2} }$

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{8^{2} }$

E = - 0.77 x 10³ V/m

8 0

2 years ago

Three disks are spinning independently on the same axle without friction. Their respective rotational inertias and angular speed

ololo11 [35]

Answer:

3/7 ω

Explanation:

Initial momentum = final momentum

I(-ω) + (2I)(3ω) + (4I)(-ω/2) = (I + 2I + 4I) ωnet

-Iω + 6Iω - 2Iω = 7I ωnet

3Iω = 7I ωnet

ωnet = 3/7 ω

The final angular velocity will be 3/7 ω counterclockwise.

7 0

3 years ago

Read 2 more answers

A bag of sugar weighs 5.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixt

gizmo_the_mogwai [7]

Answer:

Earth: 22.246 N

Moon: 3.71 N

Jupiter: 58.72 N

Explanation:

The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.

To calculate anything, we need to convert to standard measurements.

5.00 lbs = 2.27 kg

On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N

Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N

Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N

3 0

3 years ago

Convert the following statement to the language used by physicists, "I am cold, please turn on the heat."

Vikki [24]

Answer:

Explanation:fog

6 0

3 years ago