Ask question

Ask question

All categories

2 years ago

6

When a cyclist rides past you on the street, this is an example of

1 answer:

Annette [7]2 years ago

7 0

Answer:

I dont know but i think it is wind energy

Explanation:

You might be interested in

A 34-kg child runs with a speed of 2.8 m/s tangential to the rim of a stationary merrygo-round. The merry-go-round has a momentu

tekilochka [14]

Answer: $0.43\ rad/s$

Explanation:

Given

Mass of child $m=34\ kg$

speed of child is $v=2.8\ m/s$

Moment of inertia of merry go round is $I=510\ kg.m^2$

radius $r=2.31\ m$

Conserving the angular momentum

$\Rightarrow mvr=I\omega \\\Rightarrow 34\times 2.8\times 2.31=510\times \omega\\\\\Rightarrow \omega=\dfrac{219.912}{510}\\\Rightarrow \omega=0.43\ rad/s$

7 0

2 years ago

How do scientists classify species?

serious [3.7K]

Answer:

D. By comparing traits

Explanation:

Because age isn't genetic, as well as names, as well as who discovered, but traits are genetic.

6 0

2 years ago

Read 2 more answers

Which of the following represents a possible magnitude for the force of static friction when Xavier applied 72.1 Newtons of forc

lana66690 [7]

The possible magnitude for the force of static friction on the stationary cart is 72.1 N.

The given parameters:

<em>Applied force on the cart, F = 72.1 N</em>

<em />

Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.

F = ma

Static frictional force is the force resisting the motion of an object at rest.

$\Sigma F = 0\\\\F -F_f = 0$

where;

$F_f$ is the frictional force

$F= F_f \\\\72.1 = F_f\\\\F_f = 72.1\ N$

Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.

Learn more about Newton's second law of motion: brainly.com/question/25307325

8 0

2 years ago

A 50-kg copper block initially at 140Â°c is dropped into an insulated tank that contains 90 l of water at 10Â°c. Determine the f

xxMikexx [17]

Answer:

$T_f=24.71$

Explanation:

From the question we are told that:

Mass of block $m=50$

Temperature of block $T_b =140 \textdegree C$

Volume of water $V= 90L$

Temperature of water $T_w=10 \textdegree C$

Density of water $\rho=1000kg/m^3$

Specific heat of water $C_w=4.18KJ/kg-k$

Specific heat of copper $C_p=0.96KJ/kg-k$

Generally the equation for equilibrium stage is mathematically given by

$mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)$

$50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)$

$48(140-T_f)=376.2(T_f-10)$

$140-T_f=7.8375(T_f-10)$

$140-T_f=7.8375T_f-78.375$

$-8.8375T_f=-218.375$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=24.71$

6 0

2 years ago

which religious group did not expand their membership by great numbers during the Second Great Awakening

Dimas [21]

Answer:

The Catholics.

Explanation:

see answer.

4 0

2 years ago