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pentagon [3]
3 years ago
13

A stone is thrown with an initial speed of 11.5 m/s at an angle of 50.0 above the horizontal from the top of a 30.0-m-tall build

ing. Assume air resistance is negligible, and g = 9.8 m/s2. What is the magnitude of the horizontal displacement of the rock?
Physics
1 answer:
rewona [7]3 years ago
8 0

Answer:

The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Explanation:

Given that,

Initial speed = 11.5 m/s

Angle = 50.0

Height = 30.0 m

We need to calculate the horizontal displacement of the rock

Using formula of horizontal component

v_{x}=u\cos\theta

Put the value into the formula

v_{x}=11.5\times\cos50

v_{x}=7.39\ m/s

Hence, The magnitude of the horizontal displacement of the rock is 7.39 m/s.

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The nutritional calorie (Calorie) is equivalent to 1 kilocalorie. One pound of body fat is equivalent to about 4.10 × 103 Calori
FrozenT [24]

Answer:

Explanation:

Using

4.01 × 10^3 * 4.186 = 1.72×10^4j

In KJ 17.2kj

6 0
3 years ago
A person tries to lift each object with a force of 25 N, upward. Which
Vlad1618 [11]

Answer:

Option C. Objects 1 and 3 will not move, and objects 2 and 4 will accelerate

upward.

Explanation:

The following data were obtained from the question:

OBJECT >>>>>>>>> WEIGHT (N)

1 >>>>>>>>>>>>>>>> 35

2 >>>>>>>>>>>>>>>> 23

3 >>>>>>>>>>>>>>>> 26

4 >>>>>>>>>>>>>>>> 18

Force (F) applied = 25 N

From the above, the force applied to each object is 25N. Thus the following can be concluded based on the data given above:

For object 1:

Weight = 35 N

Force applied = 25 N

Thus, the object will not move since the weight of the object is greater than the force applied

For object 2:

Weight = 23 N

Force applied = 25 N

Thus, the object will move since the force applied is greater than the weight of the object.

For object 3:

Weight = 26 N

Force applied = 25 N

Thus, the object will not move since the weight of the object is greater than the force applied.

For object 4:

Weight = 18 N

Force applied = 25 N

Thus, the object will move since the force applied is greater than the weight of the object.

From the above illustrations, Object 1 and 3 will not move, and objects 2 and 4 will accelerate i.e move

4 0
3 years ago
Read 2 more answers
A measurement has high<br> when it is very close to the<br> true value?
jarptica [38.1K]

Answer:

Accuracy

Explanation:

Accuracy means making measurements that are close to the value precision means making measurement that are close in value to eachother but not necessarily close to the true value.

I hope this helps! If not sorry.

7 0
3 years ago
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
The heat capacity of nickel is 0.444 J/(g · °C). Calculate the amount of heat needed to raise the temperature of 18 g of nickel
azamat

The final speed of the nickel at the given quantity of heat is determined as 202.1  m/s.

<h3>Final speed of the nickel</h3>

Apply the principle of conservation of energy.

Q = mcΔθ

Q = (18)(0.444)(66 - 20)

Q = 367.63 J

Q = K.E = ¹/₂mv²

2K.E = mv²

v = √(2K.E/m)

where;

  • v is the final speed

v = √(2 x 367.63)/(0.018))

v = 202.1 m/s

Learn more about speed here: brainly.com/question/4931057

#SPJ1

5 0
1 year ago
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