What is the pressure exerted by a block of wood of mass 10 kg with an area of 0.1 m2

A paleontologist estimates that when a particular rock formed, it contained 12 mg of the radioactive isotope potassium-40, which

leva [86]

Answer:

$t = 2.52 billion \:years$

Explanation:

As we know by radioactivity law

$N = N_o e^{-\lambda t}$

so here we will have

$N = 3 mg$

$N_o = 12 mg$

now we will have

$3 = 12 e^{-\lambda t}$

$\lambda t = ln 4$

now we also know that

$\lambda = \frac{ln2}{1.26 \times 10^6 yrs}$

$t = 1.26\times 10^6\times \frac{ln4}{ln2}$

$t = 2.52 billion \:years$

7 0

2 years ago

Q: Riri wants to bake a cake. She adds flour, sugar, egg, baking soda, and yeast into a bowl and mixed them together. After all

babunello [35]

Answer:

I don't know what you mean about which changes occurred in this process but if its why the dough starts rising then its caused by the carbon dioxide in baking soda and yeast which is a fungus

8 0

2 years ago

Read 2 more answers

A car maintains a constant speed v as it traverses the hill and valley as shown below. Both the hill and valley have a radius of

Zigmanuir [339]

Answer:

As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.

(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.

(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.

(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.

(iv) At point A, the driver will feel the lightest.

(v)The car can go that much fast without losing contact with the road at A can be determined as follow:

Fn=0 - lose contact with road

Fg= mv²/r

mg=mv²/r

v=sqrt (gr)

8 0

2 years ago

Compare the magnitude of the magnetic field at the center of a circular current loop of radius 30 mm with the magnitude of the m

NISA [10]

Answer:

Ratio of magnetic field will be $\frac{B}{B'}=0.0055$

Explanation:

We have given radius of the loop r = 30 mm = 0.03 m

We know that magnetic field at the center of the loop is given by

$B=\frac{\mu _0i}{2r}$ ---------eqn 1

Number of turns in the solenoid is given as n = 3 turn per mm = 3000 turn per meter

We know that magnetic field due to solenoid is given by

$B'=n\mu _0i$ -------------eqn 2

Now dividing eqn 1 by eqn 2

$\frac{B}{B'}=\frac{\frac{\mu _0i}{2r}}{\mu _0ni}=\frac{1}{2nr}=\frac{1}{2\times 3000\times 0.03}=0.0055$

4 0

3 years ago

A parallel plate capacitor is charged by connecting it to a battery. After the steady state is reached, the electric energy stor

Vladimir79 [104]

Answer: C. The amount of work needed to charge the capacitor is UE, because when integrating the equation W = integral qdV with the correct limits yield the equation for the energy stored on a capacitor, UE = 1/2qV.

Explanation:

The claim about the amount of work that is needed to charge the capacitor and give evidence to support this claim is option C "The amount of work needed to charge the capacitor is UE, because when integrating the equation W = integral qdV with the correct limits yield the equation for the energy stored on a capacitor, UE = 1/2qV".

Option C is the correct answer because when we a capacitor is being charged, the amount of work that's being stored as a potential energy.

3 0

2 years ago