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love history [14]
2 years ago
9

Write a hypothesis about the effect of temperature and surface area on the rate of chemical reactions using this format: “If . .

. then . . . because. . . .” Be sure to answer the lesson question, “How do the factors of temperature and surface area affect the rate of chemical reactions?”
Physics
2 answers:
Paha777 [63]2 years ago
6 0

Answer:

If temperature and surface area increase, then the time it takes for sodium bicarbonate to completely dissolve will decrease, because increasing both factors increases the rate of a chemical reaction.

Explanation:

grandymaker [24]2 years ago
3 0
Effect of temperature.

"If the temperature of the substance is increased then the rate of chemical reaction is also increased because the kinetic energy is greater."

Effect of surface area.

"If the surface area is increased then the rate of reaction is increased because there will be more active sites for the reaction to occur. 
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The spectrum of Star A has an absorption line of hydrogen at 660.0 nm. The spectrum of Star B has an absorption line at 666 nm.
rjkz [21]

Answer:

The stars are moving away from us.

Explanation:

The observed wavelengths of hydrogen transition for stars A and B (660.0 nm and 666 nm respectively) are greater than that observed in the laboratory (656.2 nm). The observed long wavelengths for the stars means that the light from the stars is red-shifted.

According to the Doppler effect, red-shifted light means that the source is moving a way from the observer; therefore, we arrive at the conclusion that the stars A and B are moving away from us.

6 0
1 year ago
What type of roadway has the highest number of hazards per mile?
Oliga [24]
The roadway with the highest number of hazards is <span>city streets</span>
4 0
1 year ago
How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?
noname [10]

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

U=mgh

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

U=(50 kg)(9.8 m/s^2)(15 m)=7350 J

3 0
2 years ago
An insurance company hired your group to help investigate an insurance claim following a car accident. In the accident, two cars
Romashka-Z-Leto [24]

Answer:

v=8m/s

Explanation:

To solve this problem we have to take into account, that the work done by the friction force, after the collision must equal the kinetic energy of both two cars just after the collision. Hence we have

W_{f}=E_{k}\\W_{f}=\mu N=\mu(m_1+m_1)g\\E_{k}=\frac{1}{2}[m_1+m_2]v^2

where

mu: coefficient of kinetic friction

g: gravitational acceleration

We can calculate the speed of the cars after the collision by using

W_f=(0.7)(1650kg+1900kg)(9.8\frac{m}{s^2})=24353J\\24353J=\frac{1}{2}(1650kg+1900kg)v^2\\v=\sqrt{\frac{24353J}{1775kg}}=3.70\frac{m}{s}

Now , we can compute the speed of the second car by taking into account the conservation of the momentum

P_b=P_a\\m_1v_1+m_2v_2=(m_1+m_2)v\\\\v_2=\frac{(m_1+m_2)v-m_1v_1}{m_1}\\\\v_2=\frac{(1650kg+1900kg)(3.7\frac{m}{s})-(1650kg)(16\frac{m}{s})}{(1650kg)}\\\\v_2=8\frac{m}{s}

the car did not exceed the speed limit

Hope this helps!!

7 0
2 years ago
Read 2 more answers
What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56×104v/m and a magnetic fiel
sashaice [31]

1) 3.38\cdot 10^6 m/s

When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude

F_E = F_B\\qE = qvB

where

q is the electron charge

E is the magnitude of the electric field

v is the electron speed

B is the magnitude of the magnetic field

Solving the formula for v, we find

v=\frac{E}{B}=\frac{1.56\cdot 10^4 V/m}{4.62\cdot 10^{-3} T}=3.38\cdot 10^6 m/s

2) 4.1 mm

When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:

qvB=m\frac{v^2}{r}

where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find

r=\frac{mv}{qB}=\frac{(9.1 \cdot 10^{-31} kg)(3.38\cdot 10^6 m/s)}{(1.6\cdot 10^{-19} C)(4.62\cdot 10^{-3}T)}=4.2\cdot 10^{-3} m=4.1 mm

3) 7.6\cdot 10^{-9}s

The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, 2 \pi r, and the period, T:

v=\frac{2\pi r}{T}

Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:

T=\frac{2\pi r}{v}=\frac{2\pi (4.1\cdot 10^{-3} m)}{3.38\cdot 10^6 m/s}=7.6\cdot 10^{-9}s

7 0
2 years ago
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