Answer:
23.96 N
Explanation:
From the question given above, the following data were obtained:
Mass of Chihuahua (m) = 3.63 kg
Velocity (v) = 3.3m/s
Time (t) = 0.50 s
Force (F) =?
Next, we shall determine the acceleration of the Chihuahua. This can be obtained as follow:
Velocity (v) = 3.3m/s
Time (t) = 0.50 s
Acceleration (a) =?
a = v/t
a = 3.3/0.5
a = 6.6 m/s²
Thus, the acceleration of the Chihuahua is 6.6 m/s².
Finally, we shall determine the force need to stop the Chihuahua as shown below:
Mass of Chihuahua (m) = 3.63 kg
Acceleration (a) = 6.6 m/s².
Force (F) =?
F = ma
F = 3.63 × 6.6
F = 23.96 N
Therefore, a force of 23.96 N is needed to stop the Chihuahua.
The resultant vector is 5.2 cm at a direction of 12⁰ west of north.
<h3>
Resultant of the two vectors</h3>
The resultant of the two vectors is calculated as follows;
R = a² + b² - 2ab cos(θ)
where;
- θ is the angle between the two vectors = 45° + (90 - 57) = 78⁰
- a is the first vector
- b is the second vector
R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)
R² = 27.02
R = 5.2 cm
<h3>Direction of the vector</h3>
θ = 90 - 78⁰
θ = 12⁰
Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.
Learn more about resultant vector here: brainly.com/question/28047791
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It's because of the position of tectonic plates.
Answer:
23.086 mile/h
Explanation:
Given,
Distance Tyson Gay run = 100 m
time of run, t = 9.69 s
average speed of the in mph = ?
Speed of the Gay = 

v = 10.32 m/s
1 m = 3.281 ft
10.32 m = 33.86 ft
1 mile = 5280 ft
1 ft = 1.8939 x 10⁻⁴ mile
33.86 ft/s = 6.413 x 10⁻³ miles/s
Speed of Tyson in mile/hr = 6.413 x 10⁻³ x 3600
= 23.086 mile/h
Hence, speed of Tyson Gay's in mile/ hr is equal to 23.086 mph.
Answer:
The magnitude of the acceleration is equal to 19.6m/s² and the acceleration is directed upwards though the magnitude of the charge has doubled. This is because the electric force is directed upwards and from newton's second law of motion the charge will have acceleration in the same direction as the electric force on the charge.
Explanation:
The detailed solution can be found in the attachment below.
Thank you for reading and I hope this is helpful to you.