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3 years ago

A student charges a balloon and then brings it near a metal sphere hanging from the

Physics

2 answers:

Neko [114]3 years ago

8 0

Answer: B

Explanation:

The balloon will pick up stray electrons, so it will become negatively charged. Since it is attracted to the metal sphere, the sphere must be positively charged.

Hunter-Best [27]3 years ago

3 0

Answer:

Explanation:

the balloon has a negative charge and the metal sphere has a positive charge

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A cylindrical block of mass M=50kg and height h=0.2m is hanging on a rope and is in equilibrium. Any difference in atmospheric p

agasfer [191]

Answer:

$\Delta P = 1961.4\,Pa$

Explanation:

The difference of pressure is given by gauge pressure:

$\Delta P = \rho_{w}\cdot g \cdot \Delta h$

$\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.2\,m)$

$\Delta P = 1961.4\,Pa$

8 0

4 years ago

Read 2 more answers

What do all heterotrophs have in common?

ss7ja [257]

Answer:

Explanation:

your welcome dude good luck

6 0

3 years ago

A boat floats because of which force?

prisoha [69]

B is the answer - buoyant force

4 0

4 years ago

A river has a steady speed 0.500 m/s. A student swims upstream a distance of1.00 km and swims back to the starting point. a) If

Ivahew [28]

Answer:

a) 33.6 min

b) 13.9 min

c) Intuitively, it takes longer to complete the trip when there is current because, the swimmer spends much more time swimming at the net low speed (0.7 m/s) than the time he spends swimming at higher net speed (1.7 m/s).

Explanation:

The problem deals with relative velocities.

Call Vr the speed of the river, which is equal to 0.500 m/s
Call Vs the speed of the student in still water, which is equal to 1.20 m/s
You know that when the student swims upstream, Vr and Vs are opposed and the net speed will be Vs - Vr
And when the student swims downstream, Vr adds to Vs and the net speed will be Vs + Vr.

Now, you can state the equations for each section:

distance = speed × time
upstream: distance = (Vs - Vr) × t₁ = 1,000 m
downstream: distance = (Vs + Vr) × t₂ = 1,000 m

Part a). To state the time, you substitute the known values of Vr and Vs and clear for the time in each equation:

(Vs - Vr) × t₁ = 1,000 m
(1.20 m/s - 0.500 m/s) t₁ = 1,000 m⇒ t₁ = 1,000 m / 0.70 m/s ≈ 1429 s
(1.20 m/s + 0.500 m/s) t₂ = 1,000 m ⇒ t₂ = 1,000 m / 1.7 m/s ≈ 588 s
total time = t₁ + t₂ = 1429s + 588s = 2,017s
Convert to minutes: 2,0147 s ₓ 1 min / 60s ≈ 33.6 min

Part b) In this part you assume that the complete trip is made at the velocity Vs = 1.20 m/s

time = distance / speed = 1,000 m / 1.20 m/s ≈ 833 s ≈ 13.9 min

Part c) Intuitively, it takes longer to complete the trip when there is current because the swimmer spends more time swimming at the net speed of 0.7 m/s than the time than he spends swimming at the net speed of 1.7 m/s.

6 0

4 years ago

will the value of g be affected by the radius of the earth? consider the real shape of the earth. compare the acceleration due t

sp2606 [1]

Answer:

Yes, the value of g affected by the radius.

Explanation:

The formula for the force of gravity of 2 objects is

$F_{gravity} = G\frac{m_{1}m_{2}}{r^2}$ , where m1 and m2 are the masses of the 2 objects, r is the radius, and G is the gravitational constant, which is approximately $6.67 \cdot 10^{-11}$ .

Therefore, as the radius if bigger, the force of gravity is going to be smaller exponentially.

5 0

4 years ago