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3 years ago

WRONG ANSWERS WILL BE REPORTED

1 answer:

7 0

The answer to number 9 is C, decreased muscle tension. I believe number 10 is D, prepares an organism to respond to stress.

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If a 25jg car accelerates at a speed of 100 m/s^2, what will the force of the car be ?

Tema [17]

The answer is B............

6 0

3 years ago

The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca

zlopas [31]

Answer:

The value is $I = 0.0932 ML ^2$

Explanation:

From the question we are told that

The rotational inertia about one end is $I_R = \frac{1}{3} ML^2$

The location of the axis of rotation considered is $d = 0.4 L$

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is $0.4 M$

Generally the length of the rod from the its beginning to the axis of rotation consider is

$k = 1 - 0.4 L = 0.6L$

Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is

$m = 1- 0.4 M = 0.6 M$

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

$I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2$

$I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2$

Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is

$I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2$

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

$I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2$

=> $I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]$

=> $I = 0.0932 ML ^2$

8 0

3 years ago

gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an

Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0

3 years ago

A hotdog cannon shoots a hotdog straight up in the air with an initial velocity of 30 m/s. How high will the hotdog fly?

Yuri [45]

At its peak height, the hotdog will have no vertical velocity, so that

${v_f}^2-{v_i}^2=2(-g)\Delta y\implies-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}$

$\implies y_{\mathrm{max}}=46\,\mathrm m$

7 0

3 years ago

a racing car traveling initially at 8.0 m/s accelerates uniformly at 10.0 m/s^2 for 5 seconds. How far does it travel in this ti

Pepsi [2]

The car travels a distance of

(8.0 m/s) (5 s) + 1/2 (10.0 m/s²) (5 s)² = 165 m

7 0

3 years ago