Joshua was driving to a friend’s house to study. During his trip, he started on pavement. At one point, he hit an ice patch on t
2 answers:
Answer:
b. Friction decreased when he went from pavement to ice and then increased two more times.
Explanation:
Frictional force depends on the normal force of the surface and a friction coefficient.
Since we're talking about the same car, the value of will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.
After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.
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Answer:
The spring stretched by x = 13.7 cm
Explanation:
Given data
Mass = 3 kg
k = 120
Angle = 34°
From the free body diagram
Force acting on the box = mg sin
⇒ F = 3 × 9.81 ×
⇒ F = 16.45 N ------- (1)
Since box is attached with the spring so a spring force also acts on the box.
= k x
= 120
-------- (2)
The net force acting on the body is given by
Since acceleration of the box is zero so
Put the values from equation (1) & (2) we get
16.45 = 120
x = 0.137 m
x = 13.7 cm
Therefore the spring stretched by x = 13.7 cm
Answer:
The right response will be "450 volts".
Explanation:
The given values are:
R1 = 4.00 cm
R2 = 6.00 cm
q1 = +6.00 nC
q2 = −9.00 nC
As we know,
The potential difference between the two shell's difference will be:
⇒
On substituting the values, we get
Δ
<h2>Answer:</h2>
<em>Hello, </em>
<h3><u>QUESTION)</u></h3>According to the second Newton's Law,
<em>✔ We have : F = m x a ⇔ m = F/a </em>
- m = 3000/15
- m = 200 kg
The mass of the object is therefore 200 kg.