When the current through an electromagnet ceases, what generally happens to the magnetic field?

While in a car at 4.47 meters per second a passenger drops a ball from a height of 0.70 meters above the top of a bucket how far

viktelen [127]

Answer:

1.7 m

Explanation:

$v_x$ = Velocity of ball in x direction = 4.47 m/s

$u_y$ = Velocity of ball in y direction = 0

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

t = Time taken

$s_y$ = Vertical displacement = 0.7 m

$s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}$

Horizontal displacement is given by

$s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}$

The passenger should throw the ball 1.7 m in front of the bucket.

5 0

2 years ago

Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M

stealth61 [152]

The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

- For M1, the only two forces acting on it are the force of gravity $M_1 g$ (downward) and the tension in the string T (upward). So we can write

$M_1 g - T = M_1 a$

where

$M_1$ is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$a$ is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, $M_2 g sin \theta$ . So the equation of motion is

$T-M_2 g sin \theta = M_2 a$

where

$M_2 = 13.5 kg$ is the mass of the 2nd block

$\theta=35.5^{\circ}$ is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

$a=0$

So the two equations become:

$M_1 g - T=0\\T-M_2 g sin \theta = 0$

Isolating T from the 1st equation,

$T=M_1 g$

And substituting into the 2nd equation, we can find the value of the mass $M_1$ :

$M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg$

Learn more about acceleration and forces:

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#LearnwithBrainly

7 0

2 years ago

A 6.00 g lead bullet traveling at420 m/s is stopped by a large

Westkost [7]

Answer:

The increase in temperature of the bullet is 351.1 kelvin

Explanation:

First, we should find the kinetic energy of the bullet is:

$K=\frac{mv^{2}}{2}=\frac{(6\times10^{-3}\,kg)(420\,\frac{m}{s})^{2}}{2}$

with m the mass and v the velocity.

$K=529.2 J$

Now we know that half of the kinetic energy of the bullet is transformed into internal energy, by second's law of thermodynamics that means heat (Q) to raise bullet temperature (T), so:

$Q=\frac{K}{2}= 264.6\,J$