What is the mass of an object that weigh 150 N on earth

A car starts from rest and travels for 5.8 s with a uniform acceleration of 1.6 m/s² in the negative direction. What is the fina

elena-s [515]

Answer:

Final velocity of the car will be -9.28 m/sec

Explanation:

We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec

Acceleration of the car $a=1.6m/sec^2$ in negative direction so acceleration will be $a=-1.6m/sec^2$

From first equation of motion we know that

v = u+at

So $v=0+(-1.6)\times 5.8=-9.28m/sec$

So final velocity will be -9.28 m/sec

8 0

3 years ago

A hockey puck has a mass of 0.107 kg and is at rest. A hockey player makes a shot, exerting a constant force of 28.0 N on the pu

gregori [183]

Answer:

The speed does it head toward the goal = 41.87 $\frac{m}{s}$

Explanation:

Mass = 0.107 kg

Initial velocity ( u ) = 0

Force (F) = 28 N

Time = 0.16 sec

From newton's second law, Force = mass × acceleration

⇒ F = m × a

⇒ 28 = 0.107 × a

⇒ a = 261.7 $\frac{m}{s^{2} }$ --------- (1)

This is the value of acceleration.

Final speed of the mass is calculated by the equation V = U + at

⇒ U = 0 because mass in in rest position at start.

⇒ V = a t

Put the values of acceleration and time in above formula we get

⇒ V = 261.7 × 0.16

⇒ V = 41.87 $\frac{m}{s}$

Therefore the speed does it head toward the goal = 41.87 $\frac{m}{s}$

5 0

3 years ago

Hello, I have three questions to ask! It's very important, and it would be amazing if you could help me answer it :)!

geniusboy [140]

Answer:

Explanation:

true

8 0

3 years ago

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Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa

zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, $\rho _o$ = 926 kg/m³

density of the wood, $\rho _{wood}$ = 974 kg/m³

density of water, $\rho _w$ = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

$F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm$