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Sindrei [870]
4 years ago
10

Sensitization is a result of direct contact with the abrasive material, localized, and related to the concentration of the subst

ance applied.
(A) True
(B) False
Physics
1 answer:
natka813 [3]4 years ago
5 0
True because if it wasn’t true it would be in the true mine state
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Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial veloc
svet-max [94.6K]

Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(c) t₃=1.5 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

Displacement S from t₁ to t₅

Velocity V from t₁ to t₅

Solution

According to kinematic equation of motion and given information conclude that v is given by

v=v_{i}+gt\\

Also get the equation of displacement

S=v_{i}t+(1/2)gt^{2}

These two formula are used to find velocity as well as displacement for time t₁ to t₅

For t₁=0.5 s

v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\  And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m

For t₂

v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\  And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m

For t₃

v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\  And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m

For t₄

v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\  And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m

For t₅

v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\  And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m

4 0
4 years ago
a rocket has a mass 250(10^3) slugs on earth. Specify its mass in si units and its weight in si unites. if the rocket is on the
Katena32 [7]

Answer:

W_{earth} = 35.74 * 10^6 N : Rocket weight on earth

W_{moon} = 5.91 * 10^6 N : Rocket weight on moon

Explanation:

Conceptual analysis

Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:

W = m × g Formula (1)

W: weight

m: mass

g: acceleration due to gravity

The mass of a body on the moon is equal to the mass of a body on the earth

The acceleration due to gravity on a body is different on the moon and on the earth

Equivalences

1 slug = 14.59 kg

Known data

m_{earth} = m_{moon} = 250 * 10^3 slug = 250* 10^3slug * \frac{14.59kg}{1slug} = 3647.5* 10^3 kg

g_{moon}= 1.62 \frac{m}{s^2}

g_{earth}= 9.8 \frac{m}{s^2}

Problem development

To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):

W_{earth} = 3647.5* 10^3 kg * 9.8 \frac{m}{s^2} = 35.74 * 10^6 N : Rocket weight on earth

W_{moon} = 3647.5* 10^3 kg * 1.62 \frac{m}{s^2} = 5.91 * 10^6 N : Rocket weight on moon

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3 years ago
Does a emp effect electronics it hits forever and does it fry it
Free_Kalibri [48]
I recently received the following question regarding EMP effects: “What will happen to vehicles with electronic ignitions, a Chevy with an ignition module, but they are not hooked to a battery, with no path for electricity to follow, can it do damage? There may not be power on the grid, but what about a generator? Can a drill that was not plugged in still be able to run?”

The short answer is, maybe.
7 0
3 years ago
12 g … _____________________kilograms
Julli [10]
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4 years ago
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Which of the following is a trend in the health sciences
forsale [732]

Answer

more patient-centered care / advancing technology / integrative medicine

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3 years ago
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