An earthquake causes a 3 kg book to fall from a shelf. If the book lands with

A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh

irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

$K=\frac{1}{2}mv^2$

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

$U=mgh$

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

$a=\frac{v-u}{t}$

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

$u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s$

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

$\Delta v = 0 - (+9.8 m/s)=-9.8 m/s$

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

$a=\frac{v-u}{t}$

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

$v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s$

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - 0=-9.8 m/s$

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

$\Delta v = v -u$

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s$

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

$F=mg$

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

$mg = ma$

which means that the acceleration is

$a= g = -9.8 m/s^2$

and the negative sign means it points downward.

7 0

3 years ago

A 100-N force causes an object to accelerate at 2 m/s/s. What is the mass of the object?

Kitty [74]

Answer:

$50\; \rm kg$ .

Explanation:

By Newton's Second Law, the acceleration $a$ of an object is proportional to the net force $\sum F$ on it. In particular, if the mass of the object is $m$ , then

$\sum F = m \cdot a$ .

Rewrite this equation to obtain:

$\displaystyle m = \frac{\sum F}{a}$ .

In this case, the assumption is that the $100\; \rm N$ force is the only force that is acting on the object. Hence, the net force $\sum F$ on the object would also be

Make sure that all values are in their standard units. Forces should be in Newtons (same as $\rm kg \cdot m \cdot s^{-2}$ , and the acceleration of the object should be in meters-per-second-squared ( $\rm m \cdot s^{-2}$ ). Apply the equation $\displaystyle m = \frac{\sum F}{a}$ to find the mass of the object.

$\displaystyle m = \frac{100\; \rm N}{2\; \rm m \cdot s^{-2}} = 50\; \rm kg$ .

4 0

2 years ago

A student conducts an experiment to determine how the temperature of water affects the time for sugar to dissolve. In each trial

Sveta_85 [38]

B- the student changed too many variables

7 0

2 years ago

What is the period and frequency of a water wave if 4.0 complete waves pass a fixed point in 10 seconds

olga2289 [7]

The correct answer for this question is this one: C) 2.5s. The period and frequency of a water wave if 4.0 complete waves pass a fixed point in 10 seconds is that 2.5 s

Here are the following choices:
A) 0.25s
B) 0.40s
C) 2.5s
D) 4.0s

7 0

2 years ago

The charge density of a uniformly charged disk 0.420 m in diameter is 2.92 ✕ 10−2 C/m2. What is the magnitude of the electric fi

iragen [17]

Answer:

E = {(Charge Density/2e0)*(1 - [z/(sqrt(z^2 - R^2))]}

R is radius = Diameter/2 = 0.210m.

At z = 0.2m,

Put z = 0.2m, and charge density = 2.92 x 10^-2C/m2, and constant value e0 in the equation,

E can be calculated at distance 0.2m away from the centre of the disk.

Put z = 0.3m and all other values in the equation,

E can be calculated at distance 0.3m away from the centre of the disk

3 0

2 years ago