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3 years ago

Help pleaseeee my brainnn stopped working once again lol

Physics

2 answers:

Andrew [12]3 years ago

6 0

Answer:

1c 2a 3b

Explanation:

julia-pushkina [17]3 years ago

3 0

1c 2a 3b for the answers ———

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A girl throws a ball of mass 0.8 kg against a wall. The ball strikes the wall horizontally with a speed of 11 m/s, and it bounce

Karolina [17]

Answer:

F = 352 N

Explanation:

we know that:

F*t = ΔP

so:

F*t = M $V_f$ -M $V_i$

where F is the force excerted by the wall, t is the time, M the mass of the ball, $V_f$ the final velocity of the ball and $V_i$ the initial velocity.

Replacing values, we get:

F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)

solving for F:

F = 352 N

3 0

3 years ago

Which tuning forks will give rise to a beat frequency of 20 Hz when sounded together with a 240 Hz tuning fork?

Lubov Fominskaja [6]

Use the concept of beat frequency to find the applicable final freqeuncy for 20Hz beat frequency.

Beat can be defined as 'the interference pattern between two sounds of slightly different frequencies0

The expression for beat frequency is given as

$f_{beat} = |f_1-f_2|$

Where,

$f_2 =$ Final frequency

$f_1 =$ Initial frequency

The beat frequency for us is 25Hz and the initial frequency is 240Hz, then

$20= |f_2-240|$

Being an absolute value, two values are possible, both in addition and subtraction:

$f_2 = 240 \pm 20$

The two possible values are

$f_2 = 220Hz$

$f_2 = 260Hz$

3 0

3 years ago

A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave

Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction $\mu =0.4$

Consider a small element at an angle \theta having an angle of $d\theta$

Normal Force $=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}$

$N=T\cdot d\theta$

Friction $f=\mu \times Normal\ Reaction$

$f=\mu \cdot N$

and $T+dT-T=f=\mu Td\theta$

$dT=\mu Td\theta$

$\frac{dT}{T}=\mu d\theta$

$\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta$

$\frac{T_2}{T_1}=e^{\mu \pi}$

$\frac{T_2}{T_1}=e^{0.4\times \pi }$

$\frac{T_2}{T_1}==e^{1.256}$

$\frac{T_2}{T_1}=3.51$

7 0

3 years ago

50 points Answer fast please

erma4kov [3.2K]

Answer:

22 hours

Explanation:

6 0

3 years ago

The end diastolic volume of a heart is 140 mL Assume that it is a sphere. At end diastole, the intraventricular pressure is 7mmI

Vera_Pavlovna [14]

Answer:

Explanation:

We know that, V = 140 mL = 0.00014 m3

Assume that it is a sphere. so, we have

V = (4/3) \pir3

r3 = (0.00014 m3) (3) / (4) (3.14)

r = \sqrt[3]{}\sqrt[3]{}3\sqrt{}3.34 x 10-5 m3

r = 1.93 x 10-7 m

(a) The wall tension at end diastole will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end diastole = 7 mmHg = 933.2 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (933.2 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 8.18 x 10-3 N

(b) The wall tension at the end of isovolumetric contraction will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end of isovolumetric contraction = 80 mmHg = 10665.7 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (10665.7 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 9.35 x 10-2 N

(d) The wall stress from A and B which will be given as :

we know that, \sigma = T / w

For part A, we have

\sigmaA = (8.18 x 10-3 N) / (0.011 m)

\sigmaA = 0.743 N/m

For part B, we have

\sigmaB = (9.35 x 10-2 N) / (0.011 m)

\sigmaB = 8.5 N/m

4 0

3 years ago