Ca+ is an example of a(n)

A pitcher claims he can throw a 0.146-kg baseball with as much momentum as a 2.70-g bullet moving with a speed of 1.50 ✕ 103 m/s

Nataly_w [17]

(a) The pitcher must throw the ball at 27.7 m/s

The momentum of an object is given by:

$p=mv$

where

m is the mass of the object

v is the object's velocity

Let's calculate the momentum of the bullet, which has a mass of

m = 2.70 g = 0.0027 kg

and a velocity of

$v=1.50\cdot 10^3 m/s$

Its momentum is:

$p=mv=(0.0027 kg)(1.50\cdot 10^{3} m/s)=4.05 kg m/s$

The pitcher must throw the baseball with this same momentum. The mass of the ball is

m = 0.146 kg

So the velocity of the ball must be

$v=\frac{p}{m}=\frac{4.05 kg m/s}{0.146 kg}=27.7 m/s$

So, the pitcher must throw the ball at 27.7 m/s.

(b) a. The bullet has greater kinetic energy

The kinetic energy of an object is given by

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

where m is the mass of the object and v is its speed.

For the bullet, we have:

$K=\frac{1}{2}(0.0027 kg)(1.50\cdot 10^3 m/s)^2=3037.5 J$

For the ball:

$K=\frac{1}{2}(0.146 kg)(27.7 m/s)^2=56.0 J$

So, the bullet has greater kinetic energy.

6 0

3 years ago

A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by V =[5.00m/s

beks73 [17]

Equation of velocity is given as

$V =[5.00m/s−(0.0180m/s^3)t^2]i^ + [2.00m/s+(0.550m/s^2)t]j^$

at t = 7.93 s

$v = 3.87 \hat i + 6.36 \hat j$

so the magnitude of the velocity is given as

$v = \sqrt{3.87^2 + 6.36^2}$

$v = 7.45 m/s$

Part b)

the direction of the velocity is given as

$\theta = tan^{-1}\frac{6.36}{3.87}$

$\theta = 58.7 degree$

part c)

for acceleration we know that

$a = \frac{dv}{dt}$

$a = -0.036 t\hat i + 0.550\hat j$

at t = 7.93 s

$a = -0.285\hat i + 0.550\hat j$

magnitude is given as

$a = \sqrt{0.285^2 + 0.550^2}$

$a = 0.62 m/s^2$

Part d)

for the direction of the motion

$\theta = tan^{-1}\frac{0.550}{-0.285}$

$\theta = 117.4 degree$

4 0

3 years ago

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

<u>Explanation:</u>

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

4 years ago

Is the temperature of a normal healthy old man is less than that of a normal healthy young man?

Inessa [10]

Answer:

The correct answer is no.

Explanation:

The temperature of a person changes depending on their age, the physical activity they are doing and until the time of day.

There is an established average temperature that is 98.6 ° F (37 ° C) as a healthy temperature.

If the temperature exceeds 100.4 ° F (38 ° C) it means that you may be ill or have an infection.

The difference in temperature by age is very minimal, but it is still a difference. Let's see:

Babies and children: from 97.9 ° F (36.6 ° C) to 99 ° F (37.2 ° C).

Adults: from 97 ° F (36.1 ° C) to 99 ° F (37.2 ° C).

Adults over age 65: 98.6 ° F (36.2 ° C).

8 0

4 years ago

What is a vector?

nalin [4]

Answer:

C. A quantity with magnitude and a direction.

Explanation:

A vector can be defined as a quantity with magnitude and direction. Some examples of vector quantities are velocity, position, displacement, force, torque, acceleration.

For example, given the following data;

Time, t = 18.5secs

Final velocity = 78m/s

Initial velocity = 0

Substituting into the equation;

$a = \frac{78 - 0}{18.5}$

$a = \frac{78}{18.5}$

Acceleration, a = 4.22m/s²

Therefore, the acceleration of the object is 4.22m/s² due North.

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

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

Where,

a is acceleration measured in $ms^{-2}$

v and u is final and initial velocity respectively, measured in $ms^{-1}$

t is time measured in seconds.

3 0

3 years ago