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2 years ago

A spring stretches 2 meters when a 12 N force is acting on it. What is the spring constant (k)? 10 N/m 0.17 N/m 2 N/m 6 N/m

Physics

1 answer:

kogti [31]2 years ago

4 0

Answer:

Explanation:

Hooke's Law is linear; mathematically, it looks like this:

F = -kΔx where k is the spring constant. Filling in and solving for k:

12 = -k(-2) so

k = 6N/m

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A test pilot flies with an acceleration of of 5 g . what is the acceleration in meter per second

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I mean if he flies 5g that means that's his average speed too

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3 years ago

your neighbour is throttling his recent bought motorbike to show off. the sound intensity measured at your window 16m away is 0.

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3 years ago

An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f

vladimir1956 [14]

Answer:

V = 0.0806 m/s

Explanation:

given data

mass quarterback = 80 kg

mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

M(quarterback) × V = m(football) × v (football) ........................1

put here value we get

80 × V = 0.43 × 15

V = 0.0806 m/s

5 0

3 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

3 years ago

Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,

dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

La magnitud de la fuerza es 20 N

b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

$\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}$

Lo que da;

$\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m$

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

$\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}$

Lo que da;

$\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N$

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

∴ Ancho = 20 N × 5 m = 100 N · m = 100 J

c) La dirección de la fuerza viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right ) = 38.9^{\circ}$

La dirección del plano viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right ) = 38.9^{\circ}$

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

6 0

3 years ago