Alex drives 1900 meters in 200 s. Assuming he does not speed up or slow down, what is his speed?

Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela

NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

x-component of the tension in the rope on the right is approximately 91.93 N

y-component of the tension in the rope on the right is approximately 71.135 N

x-component of the tension in the rope on the left is -80.0 N

y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately 11.93 N

3. The net acceleration of the wagon in the horizontal direction is approximately 0.1193 m/s².

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The x and y component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ 91.93 N

y-component = 120.0 N × sin(40.0°) ≈ 77.135 N

The x and y component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = -80.0 N

y-component = 80.0 N × sin(180°) = 0.0 N

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = 11.93 N

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

$\displaystyle a_x = \frac{F_x}{m} \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}$

The acceleration of the wagon in the horizontal direction, aₓ ≈ 0.1193 m/s².

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skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular

laiz [17]

Answer:

$L=11.3\ kg-m^2/s$

Explanation:

Given that,

Angular speed of a skater, $\omega=3\ rot/s=18.84\ rad/s$

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

$L=I\omega$

Substitute all the values,

$L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s$

So, its angular momentum is equal to $11.3\ kg-m^2/s$ .

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When 1 kg of water and 1 kg of wood absorb the same amount of heat, the change in temperature of the wood is greater than the ch

Strike441 [17]

Intermolecular forces in water are greater than those in wood. APEX

6 0

3 years ago

A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su

madreJ [45]

maximum static friction acting on the object will be

$F_s = \mu_s mg$

plug in all values

$F_s = 0.40 \times 15 \times 9.8 = 58.8 N$

So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force

So here it is given that applied force is 20 N

so here object will not move due to this force and it will remain at rest always

due to this applied force

6 0

3 years ago

Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin

ELEN [110]

C is correct. The work-force relation is given by W=F·d, where F is force vector, and d is the displacement vector. The dot is the dot product, which is a measure of how parallel the two vectors are. It can be restated as the product of two vector magnitudes times the cosine of the angle between them. Therefore work is a scalar, not a vector, since the dot product returns a scalar.
$W=Fdcos(\theta)$

3 0

3 years ago