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3 years ago

If a spring has a spring constant of 1.00 × 10^3 N/m, what is the restoring force when the mass has been displaced 20.0 cm?

1 answer:

3 0

The spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.

Explanation:

When a spring is stretched or compressed its length changes by an amount x from its equilibrium length then the restoring force is exerted.

spring constant is k = 1.00 * 10^3 N/m

mass is x = 20.0 cm

According to Hooke's law, To find restoring force,

F = - kx

= - 1.00 *10 ^3 * 20.0

F = 20000 N/m

Thus, the spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.

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The two equal strong kids are having a tug a war. What do you expect to happen to the ball in this situation

borishaifa [10]

Answer:as per as Newtons second law, The forces exerted on the rope create tension.

As such,The tension is equal to the applied force.The tension is trasmitted to the opposite side and of the rope delivering the applied force.

Hope this helps.. :)

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3 years ago

A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the

Ierofanga [76]

Answer:

The discharge rate is $Q = 0.0192 \ m^3 /s$

Explanation:

From the question we are told that

The diameter is $d = 60 \ mm = 0.06 \ m$

The head is $h = 6 \ m$

The coefficient of contraction is $Cc = 0.68$

The coefficient of velocity is $Cv = 0.92$

The radius is mathematically evaluated as

$r = \frac{d}{2}$

substituting values

$r = \frac{ 0.06 }{2}$

$r = 0.03 \ m$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.03)^2$

$A = 0.00283 \ m^2$

The discharge rate is mathematically represented as

$Q = Cv *Cc * A * \sqrt{ 2 * g * h}$

substituting values

$Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}$

$Q = 0.0192 \ m^3 /s$

6 0

3 years ago

Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of 1.30kW/m21.30⁢kW/m2. How long does it take

vivado [14]

Answer:

T=1.384×10⁶seconds

Explanation:

Given data

p (Intensity)=1.30 kw/m²

E (Energy)=1.8×10⁹ J

A (Area)=1.00 m²

T (Time required)=?

Solution

E=PT ................eq(i)

where E is energy

P is radiation power

T is time

Radiating Power is given as

P=pA

Where p is intensity

A is Area

Put P=pA in eq(i) we get

E=pAT

T=E/pA

$T=\frac{(1.8*10^{9}}{(1.30*10^{3} )*(1.00)} )\\T=1.38*10^{6} seconds$

3 0

3 years ago

(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressur

GarryVolchara [31]

Answer:

(a) The ratio of the pressure amplitude of the waves is 43.21

(b) The ratio of the intensities of the waves is 0.000535

Explanation:

Given;

density of gas, $\rho _g$ = 2.27 kg/m³

density of liquid, $\rho _l$ = 972 kg/m³

speed of sound in gas, $C_g$ = 376 m/s

speed of sound in liquid, $C_l$ = 1640 m/s

The of the sound wave is given by;

$I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}$

Where;

$P_o$ is the pressure amplitude

$P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21$

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

$I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535$

3 0

3 years ago

Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col

Ratling [72]

In an inelastic collision, only momentum is conserved, while energy is not conserved.

1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
$p_f=(m+M)v_f=4.8 kg m/s$
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find $v_f$ , the velocity of the nail and the block after the collision:
$v_f= \frac{p_f}{m+M}= \frac{4.8 kg m/s}{0.1 kg+10 kg}= 0.48 m/s$

2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
$p_i = mv_i = p_f$
Rearranging the formula, we can find $v_i$ , the velocity of the nail before the collision:
$v_i = \frac{p_f}{m}= \frac{4.8 kg m/s}{0.1 kg}=48 m/s$

6 0

3 years ago

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