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3 years ago

Which term describes when situations, events, or people make demands on your body and mind

Physics

2 answers:

pishuonlain [190]3 years ago

7 0

stress is the correct answer to this problem

mars1129 [50]3 years ago

6 0

Stress

Stress is a condition of mental strain or pressure on the body and coming about because of unfavorable or demanding conditions that the person cannot handle at that moment.

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A train moves from rest to a speed of 25 m/s in 30.0 seconds. What is the acceleration?

fredd [130]

Answer:

$a = 0.83\ m/s^2$

Explanation:

Uniform Acceleration

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+a.t$

Where:

vf = Final speed

vo = Initial speed

a = Constant acceleration

t = Elapsed time

It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.

Solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting:

$\displaystyle a=\frac{25-0}{30}$

$\boxed{a = 0.83\ m/s^2}$

4 0

2 years ago

To heat 1g of water by 1 C requires A) 1 calorie b)1 Carlorie c) 1 Joule d) 1 watt

blsea [12.9K]

I think the answer would be 1 watt but i'm not sure

8 0

3 years ago

An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

7 0

3 years ago

A current of 3.75 A in a long, straight wire produces a magnetic field of 2.61 Î¼T at a certain distance from the wire. Find thi

Bad White [126]

Given :

Current, I = 3.75 A .

Magnetic Field, $B = 2.61\times 10^{-4}\ T$

To Find :

The distance from the wire.

Solution :

We know,

$B = K\dfrac{2i}{d}\\\\d = 10^{-7}\times \dfrac{2\times 3.75}{2.61\times 10^{-4}}\\\\d = 0.00287\ m \\\\d = 2.87\times 10^{-3}\ m$

Hence, this is the required solution.

5 0

2 years ago

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

$I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}$

$I =mr^2(0.287)$

I = 0.287 MR²

3 0

3 years ago