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3 years ago

What is the volume of a 1.2kg and displaced 1.0g/cm3

Physics

1 answer:

Zinaida [17]3 years ago

4 0

Mass = 1.2 kg = 1200 grams.

Volume = mass/density = 1200 cm3.

Hope this helps!

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When a manufacturer intentionally misleads customers by only revealing part of the truth about the evidence that supports a prod

torisob [31]

The answer would be A. failure to close

7 0

3 years ago

What is the average power output of an athlete who can life 9.0 * 10^2 kg 2.5 m in 2.0 s?

eduard

Answer:

Average power output of athlete = 11025 Watts

Explanation:

Work done is defined as the product of force applied and the displacement perpendicular to the force.

Work = $Force\times Displacement$

Power is defined as work done per unit time.

Power = $\frac{Work done}{Time}$

Here the person lifts 900 kg.

Height = 2.5 m

Time interval = 2 seconds

Force = $weight \times gravity$

= $900 \times 9.8$

= 8820 N

Work done = $Force\times Displacement$

= $8820 \times 2.5$

= 22050 J

Power = $\frac {22050}{2}$

= 11025 Watt

3 0

3 years ago

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, $v_{o} = 20.0 m/s$

Angle made by the ramp, $\theta = 22.0^{\circ}$

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH$

where

H = $dsin22^{\circ} = 5sin22^{\circ}$

g = $9.8 m/s^{2}$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ}}$ = 19.06 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m$

(b) now, considering the coefficient of friction bhetween ramp and the ball, $\mu = 0.150$ :

velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH - \mu gd$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5}$ = 18.7 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m$

6 0

2 years ago

What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 85 km/h so that the occ

borishaifa [10]

Answer:

k = 5178.8 N/m

Explanation:

As we know that spring mass system will oscillate at angular frequency given as

$\omega = \sqrt{\frac{k}{m}}$

now we have

$\omega = \sqrt{\frac{k}{1200}}$

now the maximum acceleration of the spring block system is at its maximum compression state which is given as

$a = \omega^2 A$

here A= maximum compression of the spring

so here in order to find maximum compression of the spring we will use energy conservation as we know that initial total kinetic energy of the car will convert into spring potential energy

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2$

here we know that

v = 85 km/h

$v = 85 \times\frac{1000}{3600} = 23.61 m/s$

now we have

$(1200)(23.61^2) = kA^2$

$A^2 = \frac{6.68 \times 10^5}{k}$

now from above equation of acceleration we have

$5.0 g = (\frac{k}{m})\sqrt{\frac{6.68 \times 10^5}{k}}$

$5.0(9.81) = \sqrt{k}(0.68)$

$k = 5178.8 N/m$

6 0

3 years ago

In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw

RSB [31]

Answer:

The coefficient of kinetic is

$u_{k}=0.59$

Explanation:

The forces in the axis 'x' and 'y' using law of Newton to find coefficient of kinetic friction

ΣF=m*a

ΣFy=W-N=0

ΣFy=Fn-Fu=m*a

$F_{u} =u_{k} *N\\F_{N}=25N\\N=W\\N=3.5kg*9.8\frac{m}{s^{2} }=34.3N$

$F_{N}-F_{u}=m*a\\F_{N}-u_{k}*N=m*a\\u_{k}*N=F_{N}-m*a\\u_{k}=\frac{F_{N}-m*a}{N}$

Now to find the coefficient can find the acceleration using equation of uniform motion accelerated

$v_{f} ^{2}=v_{o}^{2}+2*a(x_{f}-x_{o})\\x_{o}=0\\v_{o}=0\\v_{f} ^{2}=2*a*x_{f}\\a=\frac{v_{f} ^{2}}{2*a*x_{f}}\\ a=\frac{(1.53\frac{m}{s} )^{2}}{2*0.91m}\\a= 1.28 \frac{m}{s^{2} }$

So replacing the acceleration can fin the coefficient:

$u_{k}=\frac{F_{N}-m*a }{N}\\u_{k}=\frac{25N-(3.5kg*1.28\frac{m}{s^{2}} }{34.3N} \\u_{k}=0.59$

5 0

3 years ago