All categories

3 years ago

What is another term for a pull on an object? O A. Acceleration O B. Speed O c. Force O D. Velocity

Physics

2 answers:

hichkok12 [17]3 years ago

8 0

Answer: C. Force

Explanation: force is a pull or push

mihalych1998 [28]3 years ago

3 0

Answer:

Acceleration

Explanation:

You might be interested in

A stone is dropped into a deep well and hits the water after 3.41 seconds. How deep is<br> The well?

TiliK225 [7]

Answer:

57.0 m

Explanation:

Given:

v₀ = 0 m/s

t = 3.41 s

a = 9.8 m/s²

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (3.41 s) + ½ (9.8 m/s²) (3.41 s)²

Δx = 57.0 m

6 0

3 years ago

"Place ½ teaspoon of baking soda in each large container. If the baking soda is lumpy, first place it on a small plate and use t

photoshop1234 [79]

Answer:

The reaction only takes place if the reactants are in contact, then the surface exposed is very important because as the exposed surface of the reactants increases, the reaction velocity also increases. If the lumps are not broken up, a big amount of baking soda would be only available to react after the surrounding soda has already reacted.

8 0

4 years ago

We can reasonably model a 90-W incandescent lightbulb as a sphere 5.7 cm in diameter. Typically, only about 5% of the energy goe

o-na [289]

Answer:

See answer

Explanation:

Given quantities:

$\eta = 0.05\\ W=90[W]\\r=0.0285[m]$

where $\eta$ is the efficiency of the lightbulb (visible light is 5% of the total power), $W$ is the total power of the lightbulb, r is the radius of the lightbulb in meters.

Intensity is power divided by area:

$I =\frac{P}{A}$

a) Now the effective power is $\eta*W$ , therefore:

$I =\frac{\eta*W}{\pi r^2}=\frac{0.05*90}{4\pi (0.0285)^2}=440.87[W/m^2]$

b) Now the intensity is the average poynting vector is related to the magnitudes of the maximum electric field and magnetic field amplitudes, following:

$S_{average}= \frac{EB}{2\mu_{0}}[W/m]$

now $E$ and $B$ are related:

$E=cB\\ B=\frac{E}{c}$ and $c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}$

replace in $S_{average}$

$S_{average}=I= \frac{c \epsilon_{0}E^2}{2}[W/m]$

we replace the values and we get:

$E= \sqrt{\frac{2I}{\epsilon_{0}c}}$

$E = \sqrt{\frac{2(440.8)}{8.85*10^{-12}3*10^8}}=576.24[V/m]$

therefore

$B=\frac{E}{c}=\frac{576.24}{3*10^{8}}=1.92*10^{-6}[T]$

8 0

3 years ago

Which of the following is a form of kinetic energy?

aalyn [17]

Answer:

A) Sound Energy

Explanation:

Electrical and nuclear energy are examples of potential energy

6 0

4 years ago

Read 2 more answers

A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts

likoan [24]

Refer to the figure shown below.

m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g, the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where = kinetic coefficient of friction
T = the force in the hitch between the car and trailer.

Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N

Answer: The resistive force is 910 N

Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N

Answer: The force in the hitch is 1113 N

3 0

3 years ago

What is another term for a pull on an object? O A. Acceleration O B. Speed O c. Force O D. Velocity​

What is another term for a pull on an object? O A. Acceleration O B. Speed O c. Force O D. Velocity