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3 years ago

If T, = 40 N, find T, and the mass of the weight (W).

Physics

1 answer:

seraphim [82]3 years ago

5 0

Answer:4kg

Explanation:

acceleration due to gravity(g)=10m/s^2

Weight(w)=40N

Weight=mass x g

40=mass x 10

Divide both sides by 10

Mass =40/10

Mass=4kg

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The uncertainty in the position of an electron along an x axis is given as 5 x 10-12 m. What is the least uncertainty in any sim

Vsevolod [243]

Answer:

The least uncertainty in the momentum component px is 1 × 10⁻²³ kg.m.s⁻¹.

Explanation:

According to Heisenberg's uncertainty principle, the uncertainty in the position of an electron (σx) and the uncertainty in its linear momentum (σpx) are complementary variables and are related through the following expression.

σx . σpx ≥ h/4π

where,

h is the Planck´s constant

If σx = 5 × 10⁻¹²m,

5 × 10⁻¹²m . σpx ≥ 6.63 × 10⁻³⁴ kg.m².s⁻¹/4π

σpx ≥ 1 × 10⁻²³ kg.m.s⁻¹

4 0

3 years ago

Does anyone know how to solve this?

FromTheMoon [43]

Yes. There is a substantial number of people ... members of Brainly as well as non-members ... students, puzzle solvers, and just average educated thinkers, who would be able to solve it.

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2 years ago

The motion of a particle is described by the position function s(t) = 2t - 15t +33t+17,t>0 , where is measured in seconds and

8090 [49]

The time when the particle is at rest is at 1.63 s or 3.36 s.

The velocity is positive at when the time of motion is at $0$ .

The total distance traveled in the first 10 seconds is 847 m.

<h3>When is a particle at rest?</h3>

A particle is at rest when the initial velocity of the particle is zero.

The time when the particle is at rest is calculated as follows;

s(t) = 2t³ - 15t² + 33t + 17

$v = \frac{ds}{dt} = 6t^2 -30t + 33\\\\at \ rest, \ v = 0\\\\6t^2 - 30t + 33 = 0\\\\6(t- \frac{5}{2} )^2- \frac{9}{2} = 0\\\\t = 1.63\ s \ \ or \ 3.36 \ s$

The velocity is positive at when the time of motion is as follows;

$0$ .

The total distance traveled in the first 10 seconds is calculated as follows;

$2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m$

Learn more about motion of particles here: brainly.com/question/11066673

4 0

2 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

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2 years ago

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Answer:

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2 years ago