Help it’s multiple choice 11 through 15 please!

The half-life of the given isotope will be 10 days, if after 30 days only 31.25 grams are remaining from a sample of 250 grams of the sample size taken in the beginning.

The Half-life is the time which is required for a quantity to reduce the content to half of the amount present as its initial value. The term is used in nuclear physics to describe how quickly an unstable atom undergo radioactive decay or how long does stable atoms survive. The term is also used generally to characterize any type of exponential decay.

The half-life of the isotope can be calculated by the formula:

FR = 0.5n

FR = Fraction Remaining = 31.25 g / 250.0 g = 0.1250

n = number of half lives elapsed = ?

0.125 = 0.5n

log 0.125 = n log 0.5

-0.9031 = -0.3010 n

n = 3.000 half lives have elapsed

3 half lives = 30 days

1 half live = 10 days

Therefore, the half-life of the isotope will be 10 days.

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Answer:

Option C is the correct answer

Explanation:

Distance travelled by car during reaction time

[tex]=15\times0.4\\\\=6m[/tex]

The car stopped before hitting the animal by [tex]1 m[/tex]

Distance travelled during deceleration is [tex]21-6-1=14m[/tex]

Hence by [tex]v^2=u^2+2as[/tex]

We have

[tex]0^2=15^2+2 \cdot a \cdot 14\\\\a=\frac{-225}{28} \\\\=-8.03m/s^2[/tex]

Option C is the correct answer

Distance traveled during reaction time

Total distance

[tex]\\ \sf\longmapsto v^2-u^2=2as[/tex]

[tex]\\ \sf\longmapsto -(15)^2=2(14)a[/tex]

[tex]\\ \sf\longmapsto -225=28a[/tex]

[tex]\\ \sf\longmapsto a=-8.0m/s^2[/tex]

Answer:

acid rain corrodes a monument. d.

Explanation:it only makes sense because its doing something to a physical object.and the other ones aren't.

Answer:

>400N is needed to balance that lever

Answer:

The wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.

Given data in the question;

Speed of wave;

Frequency of wave;

wavelength;

To determine the wavelength of the radio wave, we use the expression for the relations between wavelength, frequency and speed.

Where  is wavelength, f is frequency and c is the speed.

We substitute our given values into the equation

Therefore, the wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.

Explanation:

Answer:

60n would be the needed force

The force needed to push a box weighing 30 N up a ramp that has an ideal mechanical advantage of 3 is equal to 10 N.

The mechanical advantage can be described as the ratio of the input force to the output force. The mechanical advantage of any machine can be determined by the ratio of the forces involved to do the work.

The ratio of the resistance force to the effort is called the actual mechanical advantage which will be comparatively less. The efficiency of a machine is always determined by equating the ratio of its output to its input.

The efficiency of the machine is equal to the ratio of the actual mechanical advantage (M.A.) and theoretical mechanical advantage. Mechanical advantage can be defined as the force produced by a machine to the force applied to it.

Given the load = 30 N and the ideal mechanical advantage = 3

Mechanical advantage = Load/ Effort

Input force or effort = Load/ M.A.

Force = 30/3

Input Force = 10 N

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Answer:

This is because some kinetic energy had been transferred to something else.

Explanation:

An inelastic collision is a collision in which there is a loss of kinetic energy. While momentum of the system is conserved in an inelastic collision, kinetic energy is not.

Answer:

the orbital in which the electron is located relative to the atom's nucleus.

Explanation:

Answer:

the principal energy level of an electron refers to the shell or orbital in which the electron is located relative to the atom's nucleus.

Answer:

The first one.

If the road becomes wet or crowded, you should ___ slow down and increase your following distance_

The safe distance between vehicles traveling in column specified by the command in light of safety requirements

The slower speed will help you save gas and avoid potential accidents. You can easily eliminate chances of rear end collision by maintaining a safe distance between your car and the vehicles ahead.

hence , a) slow down and increase your following distance is a correct option

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Answer:

So as two objects are separated from each other, the force of gravitational attraction between them also decreases. If the separation distance between two objects is doubled (increased by a factor of 2), then the force of gravitational attraction is decreased by a factor of 4 (2 raised to the second power).

Explanation:

hope his helps

Answer:

the wave is carrying more energy

Explanation:

trust me broski

Answer:

a

Explanation:

Answer:

[tex]0.6\; \rm m\cdot s^{-1}[/tex], assuming that drag of the water on the log is negligible.

Explanation:

The momentum [tex]p[/tex] of an object is equal to the product of mass [tex]m[/tex] and velocity [tex]v[/tex]. That is, [tex]p = m\, v[/tex].

If the drag of water on the log is negligible, momentum of the beaver and the log, combined, would be preserved.

The momentum of the beaver and the log, combined, was initially [tex]0\; \rm kg\cdot m \cdot s^{-1}[/tex].

The momentum of the beaver right after the dive would be [tex]7.5 \; {\rm kg} \times 4\; {\rm m \cdot s^{-1}} = 30\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

The sum of the momentum of the beaver and the log is conserved and should continue to be [tex]0\; {\rm kg\cdot m \cdot s^{-1}}[/tex] even after the dive. Since the momentum of the beaver is [tex]30\; {\rm kg \cdot m \cdot s^{-1}[/tex] after the dive, the momentum of the log should become [tex](-30)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Since the mass of the log is [tex]50\; {\rm kg}[/tex], the new velocity of this log would be:

[tex]\begin{aligned}v &= \frac{p}{m} \\ &= \frac{(-30)\; {\rm kg \cdot m \cdot s^{-1}}}{50\; {\rm kg}} \\ &= (-0.6)\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].

(The new velocity of the log is negative because the log would be moving away from the beaver.)

The speed of an object is the magnitude of velocity. For this log, a velocity of [tex](-0.6)\; {\rm m \cdot s^{-1}}[/tex] would correspond to a speed of [tex]|(-0.6)\; {\rm m \cdot s^{-1}| = 0.6\; {\rm m \cdot s^{-1}[/tex].

2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample. Option D is correct.

Mass is the quantity of the substance in the body or object. The SI unit of mass is Kilogram.

There are other units of measure,

First convert kg to gram,

Since, 1 Kg = 1000 g

2.1 kg = grams of sample

So,

Do the cross multiplication,

[tex]\rm mass\ of\ sample = \dfrac {2.1\ kg \times 1000\ g }{ 1 kg}\\\\\rm mass\ of\ sample =2100 g[/tex]

Now, convert 2100 g to nano-grams

Since, 1 g = 1 x 10⁹ ng

2100 g = ng of sample

So,

Do the cross multiplication,

[tex]\rm mass\ of\ sample = \dfrac {2100 g \times 1 \times 10^9 ng }{1\ g}\\\\\rm mass\ of\ sample = 2.1 \times 10^1^2 ng[/tex]

Therefore, 2.1 kg of  sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample.

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Answer:

papi sus

Explanation:

Answer:

Nuclear energy is the energy stored in atoms that can produce electricity.

Hope that helps. x

Hi there!

a.

The equation for centripetal acceleration is as follows:

[tex]\large\boxed{a_c = \frac{v^2}{r}}}[/tex]

Plug in the given values to solve:

[tex]\large\boxed{a_c = \frac{(65)^2}{75} = 56.33 m/s^2}[/tex]

b.

According to Newton's Second Law:

[tex]\large\boxed{\Sigma F = ma}}[/tex]

The acceleration is v²/r, so the net force is:

[tex]\large\boxed{\Sigma F = m(\frac{v^2}{r})}}[/tex]

Multiply by the given mass:

[tex]\large\boxed{\Sigma F = 700(56.33) = 39433.33 N}}[/tex]

c.

There is NO net force in the vertical direction since the object is NOT accelerating in the vertical direction (normal force and weight cancel).

Thus, the ONLY net force experienced by the object is in the horizontal direction and is EQUAL to the centripetal force.

When traveling with average speed 40 km/h, the car would cover a distance of 80 km in time t such that

40 km/h = (80 km) / t   ⇒   t = 2 h

so the total travel time is 2 h + 3 h = 5 h.

Average speed is defined as the total distance traveled divided by the time it took to cover that distance. So if the overall average speed was 30 km/h, then

30 km/h = (80 km + x) / (5 h)   ⇒   x = 70 km

where x is the distance traveled in the last 3 h of the trip.

Then the total displacement of the car is 80 km + 70 km = 150 km.

Answer:

0.9m/s^2 (yours is 0.87, so choose that)

Explanation:

formula for centripetal acceleration:

v^2/r

to find v, we know that f=1/90s, and r=180m.

v=(2pir)/T

v=(2pi(180))/90

v=12.6m/s

now plug into a=v^2/r

a=(12.6)^2/180

a=0.9m/s^2

Answer:

C

Explanation:

All of the other options are affecting the gravitational force and opposing force that pushed upwards, the diagram represents an object that is not moving up or down.

Range be R and height be h

[tex]\boxed{\sf R=\dfrac{u^2sin2\theta}{g}}[/tex]

[tex]\boxed{\sf h=\dfrac{u^2sin^2\theta}{2g}}[/tex]

u=initial velocity

theta is angle of projection.

g=acceleration due to gravity

ATQ

[tex]\\ \sf\longmapsto R=2h[/tex]

[tex]\\ \sf\longmapsto \dfrac{u^2sin2\theta}{g}=\dfrac{2u^2sin^2\theta}{2g}[/tex]

Cancelling required ones

[tex]\\ \sf\longmapsto sin^2\theta=sin2\theta[/tex]

sin2O=2sinOcosO

[tex]\\ \sf\longmapsto sin^2\theta=2sin\theta cos\theta [/tex]

[tex]\\ \sf\longmapsto \dfrac{sin^2\theta}{sin\theta cos\theta=2[/tex]

[tex]\\ \sf\longmapsto \dfrac{sin\theta}{cos\theta}=2[/tex]

[tex]\\ \sf\longmapsto tan\theta=2[/tex]

[tex]\\ \sf\longmapsto \theta=tan^{-1}(2)[/tex]

[tex]\\ \sf\longmapsto \theta=63.4°[/tex]

Done

Option B is correct

[tex]\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }} [/tex]

[tex]\qquad[/tex]______________________________

Then –

[tex]\qquad[/tex] [tex]\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }[/tex]

[tex]\qquad[/tex] [tex]\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }[/tex]

Therefore, the vertical component of velocity of projectile at this height will be–

☀️[tex]\qquad[/tex][tex] \pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }[/tex]

Answer:

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is v sintheta / √2

The fastest time trial for the first quarter checkpoint is 2.02 s.

The slowest time trial for the first quarter checkpoint is 2.7 s.

The range of the times measured for the checkpoint is 0.68 s.

The given parameters;

The fastest time trial for the first quarter checkpoint is the least measured time value.

fastest time trial = least time measured

fastest time trial =  2.02 s

The slowest time trial for the first quarter checkpoint is the highest measured time value.

slowest time trial = 2.7 s

The range of the times measured for the checkpoint is difference between the fastest time and slowest time.

Range = fastest time - slowest time

Range = 2.7 - 2.02

Range = 0.68 s

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Answer:

2.02, 2.15, 0.13

Explanation:

A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2.15, 2.05, 2.02, 2.07.

Use the table to answer the questions.

What is the fastest time trial for the first quarter checkpoint?

2.02

seconds

What is the slowest time trial for the first quarter checkpoint?

2.15 seconds

What is the range of times measured for this checkpoint?

0.13 seconds

Answer:

20 meters/sec

Explanation:

See attachment.

2m/s^2 = (v - 0)/10s

v = 20 m/s

Answer:

It can rotate in any direction. The sign of the angular acceleration depends on how you set the reference system, it can be both negative or positive.