A net force of 25.0 N causes an object to accelerate at 4.00 m/s2. What is the mass of the object?


Answer 1

Answer: 7kg I think or 6


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A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?



R\approx12.43 \,\, \Omega


We can use Ohm's Law to find the resistance R of a wire that carries a current I under a given potential difference:

V=I\,\,R\nR = (V)/(I) \nR=(87)/(7) \nR\approx12.43 \,\, \Omega


Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

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A basketball has a mass of 575 g. Moving to the right and heading downward at an angle of 31° to the vertical, it hits the floor with a speed of 4 m/s and bounces up with nearly the same speed, again moving to the right at an angle of 31° to the vertical. What was the momentum change ? (Take the axis to be to the right and the axis to be up. Express your answer in vector form.)



Taking the x axis to the right and the y axis to be up, the total change of momentum is \Delta \vec{ p} = 3.9429 (kg \ m)/(s) \hat{j}


The momentum \vec{p} is given by:

\vec{p} = m \ \vec{v}

where m is the mass and \vec{v} is the velocity. Now, taking the suffix i for the initial condition, and the suffix f for the final condition, the change in momentum will be:

\Delta \vec{ p} = \vec{p}_f - \vec{p}_i

\Delta \vec{ p} = m \ \vec{v}_f - m \ \vec{v}_i

\Delta \vec{ p} = m (\ \vec{v}_f -  \ \vec{v}_i)

As we know the mass of the ball, we just need to find the initial and final velocity.

Knowing the magnitude and direction of a vector, we can obtain the Cartesian components with the formula

\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector and θ is the angle measured from the x axis.

Taking the x axis to the right and the y axis to be up, the initial velocity will be:

\vec{v}_i = 4 (m)/(s) ( \ cos ( - (90  \ °- 31 \°)) , sin( - (90  \ ° - 31\°) ) ) =

where minus sign  appears cause the ball is going downward, and we subtracted the 31 ° as it was measured from the y axis

So, the initial velocity is

\vec{v}_i = 4 (m)/(s) ( \ cos ( - 59 \°) , sin( - 59 \°)) =

\vec{v}_i =  ( \ 2.0601  \ (m)/(s) , - 3.4286 (m)/(s)) =

The final velocity is

\vec{v}_i = 4 (m)/(s) ( \ cos ( 90  \ °- 31 \°) , sin( 90  \ ° - 31\°)) =

\vec{v}_i = 4 (m)/(s) ( \ cos ( 59 \°) , sin(  59 \°)) =

\vec{v}_i =  ( \ 2.0601  \ (m)/(s) ,  3.4286 (m)/(s)) =

So, the change in momentum will be

\Delta \vec{ p} = m (\ \vec{v}_f -  \ \vec{v}_i)

\Delta \vec{ p} = 0.575 \ kg (\  ( \ 2.0601  \ (m)/(s) ,  3.4286 (m)/(s) -  ( \ 2.0601  \ (m)/(s) , - 3.4286 (m)/(s)))

\Delta \vec{ p} = 0.575 \ kg (\  ( \ 2.0601  \ (m)/(s) -  \ 2.0601  \ (m)/(s),  3.4286 (m)/(s) +  3.4286 (m)/(s))

\Delta \vec{ p} = 0.575 \ kg (\  0 , 2 * 3.4286 (m)/(s) )

\Delta \vec{ p} = 0.575 \ kg * 2 * 3.4286 (m)/(s) \hat{j}

\Delta \vec{ p} = 3.9429 (kg \ m)/(s) \hat{j}

Lightning can sometimes occur on hot and humid summer evenings when there are no thunderstorms.(A) True
(B) False



(B) False


No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning.

B False because facts

A 24.1 N solid sphere with a radius of 0.151 m is released from rest and rolls, without slipping, 1.7 m down a ramp that is inclined at 34o above the horizon. What is the total kinetic energy of the sphere at the bottom of the ramp?What is the angular speed of the sphere at the bottom of the ramp? How many radians did the sphere rotate through as it rolled down the ramp What was the angular acceleration of the sphere as it rolled down the ramp




weight of solid sphere = 24.1 N

m = 24.1/g  =  24.1/10 = 2.41 Kg

radius = R = 0.151 m

height of the ramp = 1.7 m

angle with horizontal = 34°

acceleration due to gravity = 10 m/s²

using energy conservation

(1)/(2)I\omega^2 + (1)/(2)mv^2 = mgh

I for sphere

I = (2)/(5)mr^2         v = r ω

(1)/(2)\ (2)/(5)mr^2* (v^2)/(r^2) + (1)/(2)mv^2 = mgh

(7)/(10)mv^2 = mgh

h = (0.7 v^2)/(g)

v = \sqrt{(h * g)/(0.7)}

v = \sqrt{(1.7 * 10)/(0.7)}

v = 4.93 m/s

b) rotational kinetic energy


KE=(1)/(2)\ (2)/(5)mr^2* (v^2)/(r^2)


KE=(1)/(5)* 2.41 * 4.93^2

KE = 11.71 J

c) Translation kinetic energy


KE=(1)/(2)* 2.41 \time 4.93^2

KE=29.28\ J

Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open pneumothorax. b. empyema. c. pleural effusion. d. tension pneumothorax.



The correct answer is d. tension pneumothorax.


The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.

An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 9.40 m/s in 3.50 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?


(a) The acceleration of the bird is a = -1.02\ m/s. The negative sign indicated the opposite direction of motion. (b) The final speed is v = 11.76\ m/s.


Initial speed,u = 13\ m/s

Final speed, 9.4\ m/s

Time, t = 3.5\ s

The acceleration can be computed from the velocities and time. The standard unit of acceleration is a meter per second square.


The acceleration is computed as:

a = v-u/t\na = 9.4-13/3.5\na = -1.02\ m/s

Hence, the acceleration of the bird is a = -1.02\ m/s. The negative sign indicated the opposite direction of motion.


The final speed as the given time can be computed from the first equation of motion. The first equation of motion gives the relation between final and initial speed, acceleration, and time.

The final speed at time 1.2 seconds is equal to:

v = u+at\nv = 13+(-1.02)*1.20\nv = 11.76\ m/s

Hence, the final speed is v = 11.76\ m/s.

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