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3 years ago

Refer to the graphs to compare the rate of this reaction with and without a catalyst.

Physics

2 answers:

expeople1 [14]3 years ago

7 0

A catalyst lessens the energy for a reaction so it can speed up so the catalyst is not used up in the end of the reaction so the answer would be D.

Troyanec [42]3 years ago

5 0

D,the answer is D.i hope that helped.:)

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An elevator filled with passengers has a mass of 1.70×103kg. (a) The elevator accelerates upward from rest at a rate of 1.20m/s2

Anton [14]

Answer with Explanation:

We are given that mass of an elevator filled with passengers = $1.70\times 10^3 Kg$

a.a= $1.2 m/s^2$

t=1.50 s

Initial velocity=0

We have to calculate the tension in the cable supporting the elevator.

According to newton's second law,the net force is given by

$F_net=\sum F=ma$

$F-mg=ma$

Substitute the values then we get

$F=ma+mg=m(a+g)=1.7\times 10^3(1.2+9.8)=1.87\times 10^4 N$

Hence, the tension in the cable supporting the elevator= $1.87\times 10^4 N$

b.t=8.5 s

We have to find the tension in the table during this time.

Fnet=0

F=W= $1.7\times 10^3\times 9.8=1.67\times 10^4 N$

Hence, the tension in the cable during this time = $1.67\times 10^4 N$

c. Deceleration= $0.6 m/s^2$

t=3 s

We have to find the tension in the cable during deceleration.

$F-mg=-ma$

$F=mg-ma=m(g-a)$

$F=1.7\times 10^3(9.8-0.6)=1.56\times 10^4 N$

Hence, the tension in the cable during deceleration= $1.56\times 10^4 N$ .

d.We have to find that the height of elevator moved above its original staring point .

We have to find the final velocity.

$y_1=ut_1+\frac{1}{2}a_1t_1^2$

$y_1=0(1.50+\frac{1}{2}(1.2)(1.5)=1.35 m$

Final velocity

$v_1=a_1t_1$

$v_1=(1.2)(1.5)=1.8 m/s$

$y_2=v_1t_1=(1.8)(8.5)=15.3 m$

The third distance $y_3$ is up in the direction because the elevator decelerate upward which mean the distance is positive and is given by

$y_3=(1.8)(3)-\frac{1}{2}(0.6)(3)^2=2.7 m$

The total distance moved by elevator from its original position

$d=1.35+1.8+2.7=19.35 m$

Final velocity=0

5 0

4 years ago

The difference in potential between the accelerating plates of a TV set is about 24 kV. The distance between these plates is 1.5

ryzh [129]

Answer:

So electric field between the plates will be equal to $1600\times 10^3KN/C$

Explanation:

We have given potential difference between accelerating plates = 24 KV = 24000 volt

Distance between the plates d = 1.5 cm = 0.015 m

We know that potential difference is given by V = Ed, here E is electric field and d is distance between plates

So $24000=E\times 0.015$

E = 1600000 N/C = $1600\times 10^3KN/C$

So electric field between plates will be equal to $1600\times 10^3KN/C$

7 0

3 years ago

A + 32.2 μC charge feels a 0.544 N force from a + 12.3p μC charge. How far apart are they? (u stands for micro.) [?] m

love history [14]

Answer:

r = 2.55 m

Explanation:

Given that,

First charge, q₁ = 32.2 μC

Second charge, q₂ = + 12.3 μC

The force between charges, F = 0.544 N

We need to find the distance between charges. The force between two charges is given by the formula as follows :

$F=k\dfrac{q_1q_2}{r^2}\\\\r=\sqrt{\dfrac{kq_1q_2}{F}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 32.2\times 10^{-6}\times 12.3\times 10^{-6}}{0.544}} \\\\r=2.55\ m$

So, the charges are 2.55 m apart.

5 0

3 years ago

This question is about the alpha particle experiment

Wewaii [24]

This suggest the the atom has a very small positively charged nucleus in the mass of the atom is concentrated.
And the electrons revolves around the nucleus in their orbits.

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4 years ago

If you are an astronaut in the middle of the near side of the moon during a full moon,how would the ground around you look?How w

Radda [10]

The moon would be bright and the earth would be darker because the sun is on the opposite side of the earth at that time and the light from it is reflecting off the moon to produce light upon the nigh also.......

You wouldn’t see the sun a night...

Unless you lived in the north/south pole

3 0

3 years ago

Read 2 more answers