The resultant of the given forces is; 6√2 N
<h3>How to find the resultant of forces</h3>
We are given the forces as;
10 N along the x-axis which is +10 N in the x-direction
6 N along the y-axis which is +6N in the y-direction
4 N along the negative x-axis which is -4N
Thus;
Resultant force in the x-direction is; 10 - 4 = 6N
Resultant force in the y-direction is; 6N
Thus;
Total resultant force = √(6² + 6²)
Total resultant force = 6√2 N
Read more about finding resultant of a force at; brainly.com/question/14626208
Answer:
<h3>1.43m/s²</h3>
Explanation:
According to newtons second law.
F = mass * acceleration
If the doll has a mass of 0.2 kg, and the robot has a mass of 0.5 kg, the resulting mass will be 0.7kg
Force applied = 1N
acceleration = Force/mass
Substitute the values and get acceleration
acceleration = 1/0.7
acceleration = 1.43m/s²
Hence the magnitude of the acceleration of the robot is 1.43m/s²
C. Thick wire and cold temperature.
Explanation:
The resistance of a wire is given by: R = (ρL)/A
where ρ is the resistivity of the material, L is the length of the wire, A is the cross-sectional area of the wire.
From the formula, we see that the thicker the wire, the larger A, therefore the smaller the resistivity. so, a thick wire will have lower resistivity.
Moreover, the resistance of a wire increases with the temperature. In fact, high temperatures mean more motion of the atoms/electrons inside the wire, so more resistance to the flow of current through it. Therefore, colder temperature means lower resistance.
So, the correct option is thick wire and cold temperature.
Answer:
a) about 20.4 meters high
b) about 4.08 seconds
Explanation:
Part a)
To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.
In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:
Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:
Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)
To solve for "t" in this quadratic equation, we can factor it out as shown:
Therefore there are two possible solutions when each of the two factors equals zero:
1) t= 0 (which is not representative of our case) , and
2) the expression in parenthesis is zero:
