Ask question

Ask question

All categories

Svetradugi [14.3K]

3 years ago

14

A glass of water sits on a scale. The scale reads 4.78 N. A steel bolt weighs 2.07 N. The density of steel is 7.86 g/cm3. We tie

the bolt to a string and lower the bolt into the glass of water so that the bolt is fully submerged in the water but is not touching the bottom or side of the glass.

1 answer:

miv72 [106K]3 years ago

4 0

Answer:

The weight of water weight is 5.043 N.

Explanation:

Given that,

Force = 4.78 N

Weight of steel bolt = 2.07 N

Density of steel = 7.86 g/cm³

Suppose determine how much does the glass of water weight.

We need to calculate the volume of bolt

Using formula of density

$\rho_{b}=\dfrac{m}{V}$

$V=\dfrac{m}{\rho_{b}}$

Put the value into the formula

$V=\dfrac{2.07}{9.8\times7860}$

$V=2.6873\times10^{-5}\ m^3$

We need to calculate the force on bolt

Using buoyant force

$F=\rho gV$

Put the value into the formula

$F=1000\times9.8\times2.6873\times10^{-5}$

$F=0.263\ N$

We need to calculate the new weight of glass of water

Using formula for weight of glass of water

$W'_{w}=W_{w}+F_{b}$

Put the value into the formula

$W=4.78+0.263$

$W=5.043\ N$

Hence, The weight of water weight is 5.043 N.

You might be interested in

a plane passes over Point A with a velocity of 8,000 m/s north. Forty seconds later it passes over Point B with a velocity of 10

ValentinkaMS [17]

Acceleration = (change in velocity) / (time for the change)

Change in velocity = (ending velocity) - (starting velocity)

Change in the plane's velocity = (10,000 m/s north) - (8,000 m/s north)

Change in the plane's velocity = 2,000 m/s north

Time for the change = 40 seconds

Acceleration = (2,000 m/s north) / (40 seconds)

<em>Acceleration = 50 m/s² north </em>

8 0

3 years ago

[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the

ExtremeBDS [4]

Answer:

t = 5.89 s

Explanation:

To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.

Let us assume that the radius of the pulley ( $r_p$ ) = 0.4 m

Let the radius of the sphere (r) = 0.5 m

w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s

Tension (T) = 20 N

mass (m) = 3 kg each

$\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw$

$t = \frac{4r^2mw}{Tr_P}$

Substituting values:

$t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s$

7 0

3 years ago

Read 2 more answers

You lift a 10-kg box to a height of 1m. How much work do you do on the box when you lift it from the ground? (g= 9.81 m/s2)

Andreyy89

Answer:

98.1 Joule

Explanation:

Solution,

⇒Mass(m)=10kg

⇒Weight(F)=Mg=10×9.81=98.1N

⇒Distance(d)=1m

Now,

Work done=F×d

=98.1×1

=98.1J

5 0

2 years ago

A car of mass 500kg travelling at 60m/s has it speed reduced to 40m/s by a constant breaking force over a distance of 200m. find

uranmaximum [27]

Answer:

Ek1 = 900000 [J]

Ek1 = 400000 [J]

Explanation:

In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

$E_{k1}=\frac{1}{2}*m*v1^{2}$

where:

m = mass = 500 [kg]

v1 = 60 [m/s]

So we have:

Ek1 = 0.5*500*(60^2)

Ek1 = 900000 [J]

and:

Ek2 = 0.5*500*(40^2)

Ek2 = 400000 [J]

6 0

3 years ago

A powerful motorcycle can produce an acceleration of 3.50 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting

IrinaK [193]

Answer:1265 N

Explanation:

Given

acceleration of motorcycle $\left ( a\right )=3.5 m/s^2$

Velocity $\left ( v\right )=90 km/h\approx 25m/s$

Air friction and Friction $\left ( f\right )=425 N$

mass of the motorcycle with rider $\left ( m\right )=240 Kg$

Applying Forces on motorcycle

$F_{engine}-f=ma$

$F_{engine}=f+ma$

$F_{engine}=425+240\left ( 3.5\right )$

$F_{engine}=425+840=1265 N$

5 0

3 years ago