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3 years ago

11

How much heat is required to raise the temperature of 2.0 kg of concrete from 10C to 30C? (The specific heat of concrete is 880

j/kg-C.)
A. 22 J
B. 88 J
C. 8800 J
D. 35200 J

1 answer:

Phantasy [73]3 years ago

8 0

Answer:B

Explanation:

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The standard wave format for any wave is wave. When depicting wave in standard wave format, the direction of motion must be rota

german

Answer:

Transverse wave and Longitudinal wave and Electromagnetic wave

Explanation:

An inverted wave is a wave in which the vibrations of the particles are perpendicular to the direction of wave motion.
Longitudinal waves, on the other hand, are waves in which the vibrations of the particles are parallel to the direction of wave motion.
Electromagnetic waves are waves that do not require medium media for transmission, including radio waves, microwaves, UV lights, etc.
Most electromagnetic waves are transverse in nature.

4 0

3 years ago

Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the

Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan $m=12 kg$

Inclination $\theta =30^{\circ}$

Tension $T=29 N$

coefficient of Friction $\mu =0.18$

Resolving Forces Along x axis

$F_x=T\cos \theta -f_r$

where $f_r=friction\ Force$

$F_y=mg-N-T\sin \theta$

since there is no movement in Y direction therefore

$N=mg-T\sin \theta$

and $f_r=\mu N$

Thus $F_x=T\cos \theta -\mu N$

$F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))$

$F_x=25.114-18.558$

$F_x=6.556 N$

Work done by applied Force is equal to change to kinetic Energy

$F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2$

$6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2$

$v_f^2=\frac{6.556\times 2.7\times 2}{12}$

$v_f^2=2.95$

$v_f=1.717 m/s$

8 0

3 years ago

What is the concentration of H^ + at apH = 2 ? Mol / L What is the concentration of H^ + ions at apH = 6 ? Mol/L How many more H

Nonamiya [84]

Answer:

The concentration of hydrogen ion at pH is equal to 2 : $= [H^+]=0.01 mol/L$

The concentration of hydrogen ion at pH is equal to 6 : $[H^+]'=0.000001 mol/L$

There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

$pH=-\log [H^+]$

The hydrogen ion concentration at pH is equal to 2 = [H^+]

$2=-\log [H^+]\\$

$[H^+]=10^{-2}M= 0.01 M=0.01 mol/L$

The hydrogen ion concentration at pH is equal to 6 = [H^+]

$6=-\log [H^+]\\\\$

$[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L$

Concentration of hydrogen ion at pH is equal to 2 = $[H^+]=0.01 mol/L$

Concentration of hydrogen ion at pH is equal to 6 = $[H^+]'=0.000001 mol/L$

The difference between hydrogen ion concentration at pH 2 and pH 6 :

$= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L$

Moles of hydrogen ion in 0.009999 mol/L solution :

$=0.009999 mol/L\times 1 L=0.009999 mol$

There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

8 0

3 years ago

If runner A is running at 7.50 m/s and runner B is running at 7.90 m/s, how long will it take runner B to catch runner A if runn

yaroslaw [1]

Answer:

t= 137.5 s

Explanation:

So if we are wanting to figure out how long it takes runner B to catch runner A. we must first set the slope of each runner equal to one another

<u>Slopes:</u>

Runner A: y = 7.50x + 55

Runner B: y = 7.90 x

sooooo

7.50 x + 55 = 7.90 x

- 7.50 x - 7.50 x

55 = .40 x

55/.40 = .40 x / .40

x = 137.5 s

t= 137.5 s

7.50 * 137.5 + 55 =1086.25 m

7.90 * 137.5 = 1086.25 m

3 0

2 years ago

Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons

yarga [219]

Answer:

$q\approx 6.6\cdot 10^{13}~electrons$

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

$\displaystyle F=k\frac{q^2}{d^2}$

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

$\displaystyle q=\sqrt{\frac{F}{k}}\cdot d$

Substituting values:

$\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1$

$q = 1.05\cdot 10^{-5}~c$

This charge corresponds to a number of electrons given by the elementary charge of the electron:

$q_e=1.6 \cdot 10^{-19}~c$

Thus, the charge of any of the spheres is:

$\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}$

$\mathbf{q\approx 6.6\cdot 10^{13}~electrons}$

5 0

2 years ago