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3 years ago

A person carries a box of 100 kg. What is the weight of the box? (g= 9.8 m/s2)

Physics

1 answer:

aliina [53]3 years ago

8 0

Answer:

980 newton

Explanation:

100×9.8 = 980

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3 years ago

Read 2 more answers

A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is

pychu [463]

Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

Explanation:

Given;

initial velocity of the runner, u = 1.1 m/s

constant acceleration, a = 0.8 m/s²

time of motion, t = 2.0 s

The velocity of the runner at the end of the given time is calculate as;

$v = u + at$

where;

v is the final velocity of the runner at the end of the given time;

v = 1.1 + (0.8)(2)

v = 2.7 m/s

Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

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3 years ago

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nordsb [41]

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3 years ago

Density: Two blocks, A and B, are put in a tank of water. Block A has a density of 1.21 g/cm³. Block B has a density of 1.37 g/c

stira [4]

Answer:

Block A

Explanation:

Block A will float higher in the water compared to the second Block.

The density of water is 1g/cm³.

According to the principle of floatation "an object that floats in a liquid will displace equal amount of fluid to the weight of the object".

A body will become more submerged in water if it has more density because density is the mass per volume of body.

An object with a higher density than another will sink in the liquid of the one with lesser density.

Object A has lesser density and will float higher up and displace very little water.
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4 0

3 years ago

A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

4 0

3 years ago

A person carries a box of 100 kg. What is the weight of the box? (g= 9.8 m/s2)​

A person carries a box of 100 kg. What is the weight of the box? (g= 9.8 m/s2)