Answer:
6.8 m/s2
Explanation:
Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is
W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N
For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of
a = F/m = 13.6 / 2 = 6.8 m/s2
So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail
Answer:
All fraction of kinectic energy is lost to barrel of a spring gun of mass 1.8 kg
Explanation:
A ball of mass 0.50 kg is fired with velocity 160 m/s ...
The kinetic energy is given by 1/2mv²
Kinectic energy of the ball = 1/2 *0.5*160²
Kinectic energy = 1/4 *25600
Kinectic energy = 6400 joules.
If no energy is lost to fiction, and the ball sticks to a barrel of a spring gun of mass 1.8 kg with initial velocity zero, all kinetic energy is lost to the barrel of a spring gun of mass 1.8 kg.
Answer:
(B) False
Explanation:
No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning.
<span>11.823 cm
There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall.
The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm.
Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So
82 + 1/21 * 2 = 82 + 2/21 = 82.0952381
Now we have the following dimensions with a circle replacing the ball in the original problem.
Distance from wall to effective circle = 82.0952381 cm
Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm
Effective diameter of circle = 3.995462279 cm
And because the geometry makes similar triangles, the following ratio applies.
3.995462279/41.9047619 = X/124
Now solve for X
3.995462279/41.9047619 = X/124
124*3.995462279/41.9047619 = X
495.4373226/41.9047619 = X
11.82293611 = X
The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
