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3 years ago

How can a driver best prepare to enter sharp curves on a road way

Physics

1 answer:

yawa3891 [41]3 years ago

5 0

Slow down significantly before the curve in the road

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Fossil A is found in a rock layer above a layer containing Fossil B. Which fossil is probably older? A) Fossil A B) Fossil B C)

luda_lava [24]

Fossil B is older the lower they found the fossil the older it is

8 0

2 years ago

Read 2 more answers

A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 33.0 nC , respecti

matrenka [14]

Explanation:

Formula depicting relation between total flux and total charge Q is as follows.

$\phi = \frac{Q}{\epsilon_{o}}$ (Gauss's Law)

Putting the given values into the above formula as follows.

Q = $\phi \times \epsilon_{o}$

= $-953 Nm^{2}/C \times 8.854 \times 10^{-12}$

= $-8.4 \times 10^{-9} C$

= -8.4 nC

Therefore, when the unknown charge is q then,

-14.0 nC + 33.0 nC + q = -8.4 nC

q = -27.4 nC

Thus, we can conclude that charge on the third object is -27.4 nC.

7 0

3 years ago

If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is

zysi [14]

Answer:

$q_2=2.47\times 10^{-4}\ C$

Explanation:

The charge on one object, $q_1=9.9\times 10^{-5}\ C$

The distance between the charges, r = 0.22 m

The force between the charges, F = 4,550 N

Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

$F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C$

So, the charge on the other sphere is $2.47\times 10^{-4}\ C$ .

7 0

3 years ago

The temperature of a substance is _________

emmainna [20.7K]

The answer is B.
Temperature is just another term for the average kinetic energy of a substance.

6 0

3 years ago

Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet

Svetllana [295]

Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s

6 0

3 years ago