Help?? Pictures for each state

A skier is traveling fast down a mountain slope. The table shows data

melisa1 [442]

Answer:

The pushing force and the skier's mass

(A) is correct option.

Explanation:

Given that,

Mass = 64 kg

Weight = 608 N

Velocity = 21 m/s

Forward force = 45 N

We need to calculate the reaction force of the snow pushing on the skier

Using given data,

The reaction force of the snow pushing on the skier is equal to the pushing force and the skier's mass.

Hence, The pushing force and the skier's mass

(A) is correct option.

6 0

3 years ago

True or false total distance of amoving body can be negative

Charra [1.4K]

Answer:

False

Explanation:

the total distance of a moving body is always positive

Distance is the total distance between any two points

8 0

3 years ago

URGENT PLZ HELPPP!

Dovator [93]

λ = 3.125 m

λ = 2.83 m

<h3>Further explanation</h3>

Given

Frequency of radio : 96 MHz and 106 MHz

Required

The wavelength

Solution

Wavelength : from the crest to the crest of the next wave or the trough to the trough

Frequency (f): number of waves in one second

v = λ x f

λ = v : f

Input the value :

f = 96 MHz = 96 x 10⁶ Hz

λ = v : f

λ = 3 x 10⁸ : 96 x 10⁶

λ = 3.125 m

f = 106 MHz = 106 x 10⁶ Hz

λ = v : f

λ = 3 x 10⁸ : 106 x 10⁶

λ = 2.83 m

6 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

4 years ago

The system needs an ordinary friction-based brake to bring the train to a full stop. Explain why the magnetic brake is not very

BabaBlast [244]

Answer:

The slower the train is moving, the less are the changes of the magnetic flux, thus the eddy currents become weaker.

Explanation:

A magnetic brakes is not a very efficient way of braking when a train is moving slowly because at low speeds, the changes in the magnetic flux are very less and so it causes the eddy current to become weaker.

Let us find the drag force which is proportional to the velocity of two conducting plates.

The EMF that is induced in the eddy currents are : $E=v(B \times L)$

The force which is due to the induced magnetic field is, $F=l(L \times B)$

Therefore, $F=\frac{E}{R} \times (L \times B)$

$F=\frac{v(B \times L)}{R} \times (L \times B)$

Here, force is directly proportional to the velocity of the two conducting plates.

Therefore, we can say that when the speed of the train is low, the magnetic flux changes are less and thus the eddy currents are weaker.

6 0

3 years ago