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3 years ago

Will mark brainliest! Please help!

Physics

1 answer:

djverab [1.8K]3 years ago

3 0

Answer:

30 seconds

Explanation:

A = A02^-(t/hl)

--> ln(A/A0) = -(t/hl)ln2

solving for hl,

hl = -t x ln2 /ln(A/A0)

= -(60 min)xln2/ln(50/200)

= 0.5 min or 30 seconds

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If the bond enthalpy for a C-H bond is 413 kJ, what will happen when the C-H bond is broken?

pentagon [3]

Explanation:

Bond Enthalpy : It is defined as amount of energy required to break a the particular bond in there gaseous state. It is also known as bond energy. It units are kJ/mol.

Breaking of a bond is an Endothermic process (energy absorbed from the surroundings).

Formation of bond is an Exothermic process (energy is released to the surroundings).

If the average bond enthalpy for a C-H bond is 413 kJ/mol, When the C-H bond breaks in which energy will be required ,which will be an endothermic reaction.

8 0

3 years ago

A proton moves with a velocity of v with arrow = (4î − 6ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

nalin [4]

Answer:

F = [(6.4 × 10⁻¹⁹)î + (8.0 × 10⁻¹⁹)ĵ + (22.4 × 10⁻¹⁹)k] N

Magnitude of F = (2.466 × 10⁻¹⁸) N

Explanation:

The magnetic force, F, on a given charge, q, moving with velocity, v, in a magnetic field, B, is given as the vector product

F = qv × B

where v = (4î − 6ĵ + k) m/s

B = (î + 2ĵ − k) T

The particle is a proton, hence,

q = (1.602 × 10⁻¹⁹) C

F = qv × B = q (v × B)

(v × B) is given as (4î − 6ĵ + k) × (î + 2ĵ − k)

The cross product is evaluated as a determinant of

| î ĵ k |

|4 -6 1 |

|1 2 -1 |

î [(-6)(-1) - (2)(1)] - ĵ [(4)(-1) - (1)(1)] + k [(4)(2) - (-6)(1)]

î (6 - 2) - ĵ (-4 - 1) + k (8 + 6) = (4î + 5ĵ + 14k)

(v × B) = (4î + 5ĵ + 14k)

F = q (v × B) = (1.6 × 10⁻¹⁹) (4î + 5ĵ + 14k)

F = [(6.408 × 10⁻¹⁹)î + (8.01 × 10⁻¹⁹)ĵ + (22.428 × 10⁻¹⁹)k] N

Magnitude of F =

√[(6.408 × 10⁻¹⁹)² + (8.01 × 10⁻¹⁹)² + (22.428 × 10⁻¹⁹)²]

Magnitude of F = (2.466 × 10⁻¹⁸) N

Hope this Helps!!!

4 0

3 years ago

Read 2 more answers

The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh

Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

$v^2=v^2_o-2ax$

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

$\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}$

Next, consider that the formula for the force is:

$F=ma$

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

$F=(820kg)(0.66\frac{m}{s^2})=542.20N$

Hence, the required force to stop the car is 542.20N

4 0

1 year ago

For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci

MAXImum [283]

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

$z = \frac{V_o}{i_o}$

now we will have

$z = \frac{125}{3.20} = 39.06 ohm$

Part b)

Now we also know that

$\frac{R}{z} = cos\theta$

$\theta = 0.982 rad = 56.3 degree$

now we have

$\frac{R}{39.06} = cos56.3$

$R = 21.7 ohm$

5 0

3 years ago

When you let the air go out of a balloon the balloon moves or shoots like a rocket. Would this happen if there was no surroundin

Pie

Answer:

The balloon would still move like a rocket

Explanation:

The principle of work of this system is the Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (action), object B exerts an equal and opposite force (reaction) on object A"

In this problem, we can identify the balloon as object A and the air inside the balloon as object B. As the air goes out from the balloon, the balloon exerts a force (backward) on the air, and as a result of Newton's 3rd law, the air exerts an equal and opposite force (forward) on the balloon, making it moving forward.

This mechanism is not affected by the presence or absence of surrounding air: in fact, this mechanism also works in free space, where there is no air (and in fact, rockets also moves in space using this system, despite the absence of air).

3 0

3 years ago