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4 years ago

Ben and Dan both weigh 600 N. They are doing pull-ups

Physics

1 answer:

ivann1987 [24]4 years ago

6 0

Explanation:

Given parameters:

Weight of Ben = 600N

Weight of Dan = 600N

Distance of pull-up = 0.5m

Unknown:

Work done by each pull - up = ?

Solution:

Work done is defined as the product of the force applied to move a body through a certain distance.

Work done = Force x distance

Work done by Ben And Dan will be the same:

Work done = 600 x 0.3 = 180J

Work done by Ben = 180J

Work done by Dan = 180J

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The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grains are there in the ball

Fudgin [204]

Given :

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12.

To Find :

How many grains are there in the ball?

Solution :

Volume of ball of the ballpoint is :

$V = \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\times 3.14 \times 0.5^3}{3}\ mm^3\\\\V = 0.523\ mm^3$

Now, grain size of 12 has about 520000 grains/mm³.

Therefore, number of grains are :

$n = 520000\times 0.523\ grains\\\\n = 271960\ grains$

8 0

3 years ago

A block of wood of mass 300g and density 0.75 g/cm^3 is floating on the surace of a liquid of density 1.1 g/cm^3. What mass of l

Travka [436]

Answer:

The minimum mass of the lead for the combination to submerge is 155 g.

Explanation:

let M be the mass of the wood.

let m be the minimum mass of lead to be added for the combination to submerge.

let ρ1 be the density of the liquid.

let ρ2 be the density of the wood.

let ρ3 be the density of lead.

let g be the gravitational acceleration.

For the combination to submerge, the weight of the wood combined with the weight of lead should at least be equal to the buyant force, that is:

weight of wood and lead = buyant force

g×(M+m) = g×(ρ1)×(M/ρ2 + m/ρ3)

M+m = (ρ1)×(M/ρ2 + m/ρ3)

m - ρ1×m/ρ3 = (ρ1)×(M/ρ2) - M

m(1 - ρ1/ ρ3) = M(ρ1/ρ2 -1)

m = [M(ρ1/ρ2 -1)]/[(1 - ρ1/ ρ3)]

= [(300)(1.1/0.75 -1)]/[(1 - 1.1/ 11.3)]

= 155 g

Therefore, the minimum mass of the lead for the combination to submerge is 155 g.

4 0

4 years ago

A car weighing 15,000 is on a hydraulic lift platform measuring 10m^2. What is the area of the smaller piston if a force of 1,10

Artemon [7]

Answer:The pressure must be the same throughout the lift, so

P = F1/A1 = F2/A2 ⇒

A2 = (F2/F1)A1

F2 = 1,100 N

F1 = 15,000 N

A1 = 10 m2

Plug in the numbers and evaluate A2. The answer will come out in m2.

Explanation:

8 0

3 years ago

A supersonic airplane is flying horizontally at a speed of 2610 km/h.

Delvig [45]

Answer:

Centripetal acceleration of this aircraft: approximately $6.52\; {\rm m \cdot s^{-2}}$ .

Distance covered during the turn: approximately $63.2\; {\rm km}$ .

Time required for the turn: approximately $0.0242\; \text{hours}$ (approximately $87.2\; {\rm s}$ .)

Explanation:

Convert velocity and radius to standard units:

$\begin{aligned}v &= 2610\; {\rm km \cdot h^{-1}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &= 725\; {\rm m\cdot s^{-1}} \end{aligned}$ .

$\begin{aligned} r = 80.5\; {\rm km} \times \frac{1000\; {\rm m}}{1\; {\rm km}} = 8.05 \times 10^{4}\; {\rm m}\end{aligned}$ .

Hence, the centripetal acceleration of this aircraft:

$\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(725\; {\rm m\cdot s^{-1}})^{2} }{8.05 \times 10^{4}\; {\rm m}} \\ &\approx 6.53\; {\rm m \cdot s^{-2}}\end{aligned}$ .

The trajectory of the turn is an arc with a radius of $r = 80.5\; {\rm km}$ and a central angle of $\theta = 90^{\circ} = (\pi / 4)$ . The length of this arc would be:

$\begin{aligned} s &= r\, \theta \\ &= 80.5\; {\rm km} \times (\pi / 4) \\ &\approx 63.2\; {\rm km}\end{aligned}$ .

The time required to travel $63.2\; {\rm km}$ at a speed of $2610\; {\rm km \cdot h^{-1}}$ would be:

$\begin{aligned}t &= \frac{s}{v} \\ &\approx \frac{63.2\; {\rm km}}{2610\; {\rm km \cdot h^{-1}}} \\ &\approx 0.0242\; {\rm h} \\ &\approx 0.0242 \; {\rm h} \times \frac{3600\; {\rm s}}{1\; {\rm h}} \\ &\approx 87.2\; {\rm s} \end{aligned}$ .