Suppose there was 2 Newtons what would the mass of the cart be?

A series of bright fringes appears on the viewing screen of a Young's double-slit experiment. Suppose you move from one bright f

goblinko [34]

Ahahaha fidjdsd skssjsjsbs SJSU’s

6 0

3 years ago

Imagine that two balls, a basketball and a much larger exercise ball, are dropped from a parking garage. If both the mass and ra

pashok25 [27]

Here if we assume that there is no air friction on both balls then we can say

$F = mg$

now the acceleration is given as

$F = ma = mg$

$a = g$

so here both the balls will have same acceleration irrespective of size and mass

so we can say that to find out the time of fall of ball we can use

$y = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2y}{g}}$

now from above equation we can say that time taken to hit the ground will be same for both balls and it is irrespective of its mass and size

3 0

3 years ago

The ____ of a position-time graph represents an object’s velocity.

Marianna [84]

Answer:

The slope of a position-time graph represents an object’s velocity.

Explanation:

In a position-time graph, the values on the x-axis represent the time, while the values on the y-axis represent the position of the object.

Velocity is defined as the ratio between the displacement of an object and the time taken:

$v=\frac{\Delta s}{\Delta t}$

However, we can see that this definition corresponds to the slope of the curve in a position-time graph. In fact:

$\Delta s$ , the displacement, corresponds to the difference in position, so the difference between the values on the y-axis: $\Delta s=y_2 -y_1$

$\Delta t$ , the time interval, corresponds to the difference in times, so the difference between the values on the x-axis: $\Delta t= t_2 -t_1=x_2 -x_1$

So, the velocity is

$v=\frac{\Delta s}{\Delta t}=\frac{y_2 -y_1}{x_2 -x_1}$

which corresponds to the slope of the curve.

3 0

3 years ago

Read 2 more answers

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

Three small children of mas 20.0 kg, 24.0 kg, and 16.0 kg, respectively, hold hands, as

olya-2409 [2.1K]

Answer:

hope this helps youuuuu

................

7 0

3 years ago