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2 years ago

Which of the following most likely happens when the temperature of a chemical reaction is increased?

Physics

2 answers:

Natali5045456 [20]2 years ago

6 0

A molecules speed up because the temp is rising there for the molecules speed up

hope this helped =^-^=

Diano4ka-milaya [45]2 years ago

5 0

Molecules Speed Up because the other three options only occur when the temperature is decreased.

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An electric flux of 147 N*m^2/C passes through a flat horizontal surface that has an area of 0.824 m^2. The flux is due to a uni

Tresset [83]

Answer:231.16 N/C

Explanation:

Given

Electric Flux $=147 N-m^2/C$

Area(A) $=0.824 m^2$

Given Field point above $31.6 ^{\circ}$

Therefore angle between Area vector Electric Field =90-31.6= $58.4^{\circ}$

We know that Flux is given by

$\phi =\vec{E}\cdot \vec{A}$

$\phi =EAcos\theta$

$147=E\times 0.824\times cos(58.4)$

E=231.16 N/C

4 0

3 years ago

A 0.420 kg block of wood rests on a horizontal frictionless surface and is attached to a spring (also horizontal) with a 20.5 N/

Aneli [31]

Answer:

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Explanation:

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3 0

3 years ago

Two students, Jordan and Taylor, are talking about light. Jordan says that light behaves as a stream of particles. Taylor declar

FrozenT [24]

Answer:

"Both students are correct. When light is traveling through small openings, solids, liquids, gases, or a vacuum, it behaves like a wave. When it interacts with matter, it acts like a stream of particles."

Explanation:

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3 0

2 years ago

An ice skater of mass m = 60 kg coasts at a speed of v = 0.8 m/s past a pole. At the distance of closest approach, her center of

IrinaVladis [17]

Answer:

1.78 rad/s

1.70344 rad/s

Explanation:

v = Velocity = 0.8 m/s

m = Mass of person = 60 kg

$r_1$ = Distance between center of mass of person and pole = 0.45 m

$r_2$ = New distance between center of mass of person and pole = 0.46 m

I = Moment of inertia

Angular speed is given by

$\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s$

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

$L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s$

The new angular speed is 1.70344 rad/s

7 0

2 years ago

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s

dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

= 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2

= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

4 0

2 years ago