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2 years ago

1. Two blocks (5.0 kg and 6.0 kg) are connected through a frictionless system. The angle of

Physics

1 answer:

Paul [167]2 years ago

8 0

Answer: A) 45.5N b) 34.6N

Explanation:

Given the following :

Mass of block A(m1) = 5kg

Mass of block B (m2) = 6kg

Angle of incline (Θ) = 45°

Force pulling the down the incline = m1×a×sinΘ

Tension (T) = (m2×a) - (m2)×g

Where g = acceleration due to gravity

a = acceleration

Net force = (5×9.8×sin45°) - (6×9.8)

Net force = 34.648232 - 58.8

The net force acting on the body = 24.4N

Therefore,

Acceleration, a = Net force/ total mass

a = 24.4 / (6+5)

a = 2.22ms^-2

T = (m2 ×g) - (m2×a)

T = (6 × 9.8) - (6 × 2.22)

T = 58.8 - 13.32

T = 45.48

T = 45.5N

B) Normal reaction:

Horizontal component:

m1gCosΘ = 5 × 9.8 × cos45°

= 5 × 9.8 × 0.7071067

= 34.648232

= 34.6N

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A point on a wave completes one cycle every 5 seconds. The frequency of this wave is 1/5 seconds.

Sophie [7]

Answer:

True

Explanation:

1 cycle every 5 sec is the same when you set up the ratio as 1/5

Hope this helps!! Have a great day!! :)

7 0

2 years ago

A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current i

lisov135 [29]

Answer:

8.07 m/s, 81.7º NE.

Explanation:

The ship, due to the local ocean current, will be deviated from its original due north bearing.
In order to find the magnitude of the velocity of the ship, we need to convert a vector equation, in an algebraic one.
If we choose two axes coincident with the N-S and W-E directions, we can find the components of the velocity along these directions.
Clearly, the velocity of the ship, relative to water, is only due north, so it has no component along the W-E axis.
The local ocean current, as it is directed at an angle between both axes, has components along these axes.
These components can be found from the projections of the velocity vector along these axes, as follows:

$vocx = voc* cos 40 = 1.53 m/s * 0.766 = 1.17 m/s\\vocy = voc* sin 40 = 1.53 m/s * 0.643 = 0.98 m/s$

The component along the N-S axis (y-axis) of the velocity of the ship will be the sum of the velocity relative to water, plus the component of the ocean current along this same axis:

$vshy = vsw + vwy = 7.00 m/s + 0.98 m/s = 7.98 m/s$

The component along the W-E axis, is just the component of the local ocean current in this direction:

vshx = 1.17 m/s

We can find the magnitude of the velocity vector, applying the Pythagorean theorem, as follows:

$v = \sqrt{vshx^{2} + vshy^{2} } =\sqrt{(7.98m/s)^{2} +(1.17m/s)^{2} } =8.07 m/s$

The direction of the vector relative to the W-E axis (measured in counterclockwise direction) is given by the relative magnitude of the x and y components, as follows:

$tg \theta = \frac{vshy}{vshx} = \frac{7.98}{1.17} = 6.82 \\ \theta = tg^{-1} (6.82)\\ \theta= 81.7\deg$

The velocity of the ship, relative to Earth, is 8.07 m/s, 81.7º North of East.

7 0

2 years ago

Safety belts protect people in cars in the event of an accident because, according to Newton’s laws of motion, when an impact ca

Fed [463]

Answer:

D :)

Explanation:

6 0

2 years ago

Explain how the isotopes of an element are alike and how are they different

leva [86]

All isotopes of the same element have the same number of protons
in the nucleus of their atoms, but different numbers of neutrons.

3 0

2 years ago

Two objects attract each other gravitationally with a force of 2.5 x 10^-10N when they are 0.25 m apart. Their total mass is 4.0

In-s [12.5K]

Answer:

M = 3.9406 kg and m = 0.0594 kg

Explanation:

The gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically it is expressed as follows:

Fg = (G×M×m)/r² Formula (1)

Where:

Fg is the gravitational force (N)

G is the universal gravitation constant, G = 6.67 × 10⁻¹¹ (N×m²)/kg²

M and m are the masses of the bodies that interact (kg).

r is the distance that separates them (m).

Known Data

Fg = 2.5 × 10⁻¹⁰ N

r = 0.25 m

G = 6.67 × 10⁻¹¹ (N×m²)/kg²

Problem development

We propose 2 equations

M + m = 4kg

M = 4 - m equation (1)

We replace in formula (1)

2.5 × 10⁻¹⁰ = (6.67 × 10⁻¹¹ × M × m)/(0.25)²

2.5 × 10⁻¹⁰ × (0.25)² = (6.67 × 10⁻¹¹ × M × m)

(2.5 × 10⁻¹⁰ × (0.25)²)/(6.67 × 10⁻¹¹) = M × m

M × m = 0.234 equation (2)

We replace M = 4 - m in equation (2)

(4 - m) × m = 0.234

4m - m² = 0.234

m² - 4m + 0.234 = 0 (quadratic equation)

We apply the formula for the quadratic equation and obtain 2 values for m that meet the conditions:

m = 3.9406 kg or m = 0.0594 kg

We replace m in equation (1)

M = 4 - 3.9406 = 0.0594 kg or M = 4 - 0.0594 = 3.9406

To meet the condition that M + m must give 4 kg, one mass must be equal 3.9406 and the other must equal 0.0594, then:

M = 3.9406 kg and m = 0.0594 kg

6 0

2 years ago