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3 years ago

10What is the acceleration due to gravity if an object falls from a height of 25 m?!

Physics

1 answer:

jonny [76]3 years ago

3 0

Answer:

D. 9.81 m/s²

Explanation:

Near the surface of the earth, the acceleration due to gravity is 9.81 m/s² towards the earth's center.

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Si se aplic una fuerza de 150N en un área de de 0.4m2¿cual será la preciosa ejercida?

natali 33 [55]

Answer:

P = 375 Pa

Explanation:

The question says that,"If a force of 150N was applied in an area of 0.4m², what will be the precious exerted?"

We have,

Force, F = 150 N

Area, A = 0.4m²

We need to find the pressure exerted. We know that,

Pressure = forece/area

So,

$P=\dfrac{150\ N}{0.4\ m^2}\\\\P=375\ Pa$

So, the required pressure is equal to 375 pa.

7 0

3 years ago

Speedy Sue, que conduce a 30.0 m/s, entra a un túnel de un carril. En seguida observa una camioneta lenta 155 m adelante que se

Romashka [77]

Answer:

Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

Explanation:

Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:

Speedy Sue

$x_{S} = x_{S,o}+v_{S,o}\cdot t+\frac{1}{2}\cdot a_{S}\cdot t^{2}$

Camioneta lenta

$x_{C} = x_{C,o} +v_{C}\cdot t$

Donde:

$x_{S,o}$ , $x_{C,o}$ - Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.

$v_{S,o}$ - Velocidad inicial de Speedy Sue, medida en metros por segundo.

$v_{S}, v_{C}$ - Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.

$a_{S}$ - Deceleración de Speedy Sue, medida en metros por segundo cuadrado.

$t$ - Tiempo, medido en segundos.

Si conocemos que $x_{S} = x_{C}$ , $x_{S,o} = 0\,m$ , $x_{C,o} = 155\,m$ , $v_{S,o} = 30\,\frac{m}{s}$ , $v_{C} =-5\,\frac{m}{s}$ y $a_{S} = -2\,\frac{m}{s^{2}}$ , encontramos la siguiente función cuadrática:

$155\,m + \left(-5\,\frac{m}{s} \right)\cdot t = 0\,m+\left(30\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot (-2\,\frac{m}{s^{2}} )\cdot t^{2}$

$-t^{2}+35\cdot t-155 = 0$ (Ec. 1)

Las raíces de esta función son:

$t_{1}\approx 29.798\,s$ , $t_{2} \approx 5.201\,s$

La colisión ocurriría en la raíz positiva más pequeña, es decir:

$t \approx 5.201\,s$

Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: ( $v_{C,o} = -5\,\frac{m}{s}$ , $x_{C,o} = 155\,m$ , $t \approx 5.201\,s$ )

$x_{C} = 155\,m +\left(-5\,\frac{m}{s}\right)\cdot (5.201\,s)$

$x_{C} = 128.995\,m$

En síntesis, si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

0 0

3 years ago

A person hits a tennis ball with a mass of 0.058 kg against a wall.

Alla [95]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

$F=\dfrac{m(v-u)}{t}$

$F=\dfrac{0.058\times (11-(-11))}{2.1}$

F = 0.607 N

(b) Area of wall, $A=3\ m^2$

Let P is the average pressure on that area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{0.607\ N}{3\ m^2}$

P = 0.202 Pa

Hence, this is the required solution.

6 0

4 years ago

14. A block rests on a frictionless table on Earth. After a 20-N horizontal force is applied to the block, it accelerates at 3.9

joja [24]

<span>14. Let's find mass. We know that F = m*a, so m = F/a.m = 20/3,9 m/s^2 = 5,12 kg;We already know the mass, so acceleration will be a = F/m;a = 10/5,12 = 2,0 m/s^2 (approximately);Answer: E.
15. According to the picture given above, let's define all components:These are refer to y: $985N sin(31)= 507;788N cos(32)= 668;411N sin(53)=-328;$
And these are refer to x: $985N cos(31)= 844 788N sin(32)= -417 411N cos(53)= -247$
And finally let's calculate tan: $tan^-1 (507/844)= 30,98; tan^-1 (668/-417)= -58,02; tan^-1 (-328/-247)= 53,01$
According to these calculations we've got: Sum Fx = 179,4 and Sum Fy = 847. Then let's count magnitude: $Fsum = rad 179^2 + 847^2 = 865$
Then we've got [tex]cos^-1 (179/865) = 77.7 (approximately).
So the most approximate answer is C. 866 N at 78.1° counterclockwise to the x-axis.
16. I think that this question is incomplete because there wasn't mentioned in what end of the chain the tension should be calculated. Anyway I'll help you with both bottom and top ends. We will use the basic formula W=m*g; W(bottom) = 175*9.8 = 1715 N; W(top) = (175+12)*9.8 = 1833 N; So you should choose between B. 1830 N or D. 1720 N. But I think the most possible answer is B.
17. I am definitely sure that A diagram generates the most tension in one chain. So the answer is C. The box is held by 1 chain and have all its weight.
18. I think that each planet would move in a straight line at constant speed, because there will be a zero gravity condition and there won't any impact on the planets.
19. According to the Work-Energy Theorem we have:1/2*1200*2^2 = 2400 (J);Then let's count the average force according to the formula: Work = F*D where D is displacement.F = 2400/ 0.15 = 16000 (N)
So the answer is B. 1.6 × 10^4 N
20. If my memory serves me well, magnetism is the force that can act through empty space. It's one aspect of the combined electromagnetic force. So the answer is A.
21. According to the illustration given above, I think we should use formula F = m*g; Let's count m. m = 5 kg - 0,6 kg = 4,4 kg; Then we have everything to count force:F = 4,4 * 9,8 = 43,1 N. So the most approximate answer is D. 43 N.
22. I am definitely sure that the answer is C. on Earth at sea level. Because weight has the formula W=mg. And on the earth surface the magnitude of g is higher that everywhere so the greatest weight is on Earth at sea level.
23. We have everything to calculate the acceleration. According to Newton's second law, the formula is a = F/m;a = 20/40 = 0,5 m/s^2; So the answer is A.
24. According to the definition of action reaction forces, the answer should be: A. The forces are opposite in direction, with the force on the ball much stronger in magnitude.
25. I am pretty sure that we should use Newton’s second law of motion F = m*a. First we should find mass, using formula W=m*g => m = W/g => m = 2400/9.8= 245 kg. Then we can find F.F = m*a;F = 245*12 = 2940 N; Answer: B
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7 0

3 years ago

Read 2 more answers

PLEASEEEE HEEEEELP!!!!!

8090 [49]

Answer:

Before:

$p_{truck}=16400\ kg.m/s$