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3 years ago

8

một hòn đá ném theo phương nằm ngang với vân tốc v0=15 m/s. tính gia tốc tiếp tuyến và gia tốc pháp tuyến của hòn đá sau lúc ném

1 giây.

1 answer:

Lady_Fox [76]3 years ago

5 0

Answer:

what language is this???

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Help me please I need help?

sweet [91]

Your answer is c. homologous structures

4 0

3 years ago

A pendulum consists of a stone with mass m swinging on a string of length L and negligible mass. The stone has a speed of v0 whe

Dmitriy789 [7]

Answer:

(a) $v = \sqrt{ mv_0^2 - 2gL(1-\cos(\theta))}$

(b) $\theta = \arccos(\frac{2gL-v_0^2}{2gL})$

(c) $E_{total} = \frac{1}{2}mv_0^2$

Explanation:

(a) The total mechanical energy of the system is conserved.

$K_A + U_A = K_B + U_B\\\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}mv^2 + mgh \\h = L - L\cos(\theta) = L(1 - \cos\theta)\\\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2 - mgL(1-\cos(\theta))\\v^2 = mv_0^2 - 2gL(1-\cos(\theta))\\v = \sqrt{ mv_0^2 - 2gL(1-\cos(\theta))}$

(b) The conservation of energy states

$K_A + U_A = K_M + U_M\\\frac{1}{2}mv_0^2 + 0 = 0 + mgL(1-\cos(\theta))\\1-\cos(\theta) = \frac{v_0^2}{2gL}\\\cos(\theta) = 1-\frac{v_0^2}{2gL} = \frac{2gL-v_0^2}{2gL}\\\theta = \arccos(\frac{2gL-v_0^2}{2gL})$

(c) As explained in part (a) the total mechanical energy of the system is equal to the initial kinetic energy, since the potential energy of the system at that point is zero.

$E_{total}= K_A + U_A = K_A + 0\\E_{total} = \frac{1}{2}mv_0^2$

8 0

3 years ago

A basketball player can leap upward 0.42 m. What is the magnitude of his initial velocity at the start of the leap? Use g=9.81 m

Tatiana [17]

Farmula 2gh=vf²-vi²
2×0.42×9.81=0²-vi²
8.2404=vi²
Vi=✓8.2404
Vi=2.87m/sec

8 0

3 years ago

1. List down five everyday objects made from different materials with varying textures and colours.

Salsk061 [2.6K]

1.pencil
2.phone
3.bed
4. Keyboard
5.lamp

6 0

3 years ago

Objects 1 and 2 attract each other with a gravitational force

kolbaska11 [484]

Answer:

The new gravitational force will be of 4.5 units

Explanation:

Recall that the formula for the gravitational force between two objects of masses m1 and m2 separated by a distance d is given by:

$F_g=G\,\frac{m1*m2}{d^2}$

in our case, we are told that such gives 72 units of force:

$F_g=G\,\frac{m1*m2}{d^2} =72$

Then we change the distance between the objects to 4 times the original (4 * d), such will produce a new gravitational force Fg':

$F_g'=G\,\frac{m1*m2}{(4*d)^2} =G\,\frac{m1*m2}{16*d^2} = \frac{1}{16} *G\,\frac{m1*m2}{d^2}=\frac{1}{16} *\,72=4.5$

Therefore the new gravitational force would be of 4.5 units

7 0

3 years ago