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3 years ago

A 3.00 kg mass is traveling at an initial speed of 25.0 m/s. What is the

Physics

1 answer:

Nitella [24]3 years ago

4 0

Answer:

The magnitude of the force required to bring the mass to rest is 15 N.

Explanation:

Given;

mass, m = 3 .00 kg

initial speed of the mass, u = 25 m/s

distance traveled by the mass, d = 62.5 m

The acceleration of the mass is given as;

v² = u² + 2ad

at the maximum distance of 62.5 m, the final velocity of the mass = 0

0 = u² + 2ad

-2ad = u²

-a = u²/2d

-a = (25)² / (2 x 62.5)

-a = 5

a = -5 m/s²

the magnitude of the acceleration = 5 m/s²

Apply Newton's second law of motion;

F = ma

F = 3 x 5

F = 15 N

Therefore, the magnitude of the force required to bring the mass to rest is 15 N.

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What did the scientists deduce from the fact that the ants eyes of the desert have multiple lenses

Evgesh-ka [11]

That the pupl is smaller than the nulian hope this helped

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3 years ago

A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

3 years ago

Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p

ira [324]

Answer:

The force exerted on the $q_1$ is $F = 2.25*10^{3} \ N$

Explanation:

From the question we are told that

The area is $A = 2.34*10^{-3} \ m^2$

The magnitude of charge placed on them is $q = 7.07 * 10^{-7} C$

The charge placed between the plate is $q_1 = 6.62 *10^{-5} C$

The electric field generated around the plate is mathematically represented as

$E = \frac{q}{A \epsilon_o}$

Substituting values

$E = \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}$

$E = 34*10^{6} \ V/m$

The force exerted the charge $q_1$ is mathematically represented as

$F = q_1 * E$

Substituting values

$F = 6.62 *10^{-5} * 34*10^{6}$

$F = 2.25*10^{3} \ N$

8 0

3 years ago

A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea

viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

$L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}$ (1)

time taken = 5 years

We know that :

time = $\frac{distance}{speed}$

5 = $\frac{L''}{v}$ (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

$\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

$\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

$t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}$

Solving the above eqn:

t'' = 20.56 years

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3 years ago

What happens to energy and frequency of a wave if its wavelength increases?

Dahasolnce [82]

Answer:

Waves can be measured using wavelength and frequency. ... The distance from one crest to the next is called a wavelength (λ). The number of complete wavelengths in a given unit of time is called frequency (f). As a wavelength increases in size, its frequency and energy (E) decrease.

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3 years ago