Which object is moving faster? A

The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig

julia-pushkina [17]

Answer: They are all true

a. The tension in the rope is everywhere the same.

b. The magnitudes of the forces exerted on the two objects by the rope are the same.

c. The forces exerted on the two objects by the rope must be in opposite directions.

d. The forces exerted on the two objects by the rope must be in the direction of the rope.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)

- Cutiepatutie ☺❀❤

5 0

3 years ago

Ask Your Teacher In a choir practice room, two parallel walls are 4.00 m apart. The singers stand against the north wall. The or

igomit [66]

Answer: 4m

Explanation:

Since the angle of incidence of a plane mirror can be anything from 0 to 90°

Assuming that the place is a perfectly square 4×4m room

The incident ray would be 45° for the choir(object) at a 4m distance, this is still within the range of values.

We do not forget also, that the focal length of a plane mirror is infinity, the organist would in fact see farther than 4m if need be. And wider

5 0

3 years ago

A runner begins running at the beginning of 7th period and stops running at the end of the period. She runs at a pace of 10 km/h

MrRissso [65]

Answer:

She run for, t = 0.92 s

Explanation:

Given data,

The velocity of the runner, v = 10 km/h

The distance covered by the runner, d = 9.2 km

The relationship between the velocity, displacement and time is given by the formula,

t = d / v

Substituting the given values in the above equation,

t = 9.2 / 10

= 0.92 s

Hence, she ran for, t = 0.92 s

8 0

3 years ago

If a source of sound waves is rapidly approaching a person, the sound heard by the person appears to have

velikii [3]

Correct answer choice is:

D. A frequency higher than the original frequency.

Explanation:

This is a true case of Doppler's effect. The Doppler effect can be defined as the effect originated by a traveling source of waves in which there is a visible higher variation in pulse for observers towards what the source is progressing and a visible descending shift in rate for observers from what the source is dropping.

8 0

3 years ago

Read 2 more answers

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago