1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
damaskus [11]
2 years ago
5

A worker is pushing a crate of tools up a ramp into a truck. The crate has a mass of 120 kg and is accelerating at a rate of 1.0

5 m/s^2. Once he finishes pushing that crate up, he then takes another crate that is 3 times as massive and pushes it up the ramp with an acceleration of 0.71 m/s^2. What is the ratio of the force the worker used on the on the 120 kg crate and the more massive crate? a. 0.218 b. 0.493 c. 1.68 d. 1.07
Physics
1 answer:
Andrej [43]2 years ago
7 0

Answer:

b. 0.493

Explanation:

You might be interested in
Apilot of mass 70 kg rides a fighter jet The fighter jet moves in a vertical circle of radius 100 m at a constant
Cerrena [4.2K]

Answer:

the  force exerted by the seat on the pilot is 10766.7 N

Explanation:

The computation of the force exerted by the seat on the pilot is as follows:

F = Mg + \frac{MV^2}{R}\\\\= 70 \times 9.81  + \frac{70 \times 120^2}{100}\\\\= 10766.7 N

Hence, the  force exerted by the seat on the pilot is 10766.7 N

4 0
3 years ago
Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

Answer:

The curl is 0 \hat x -z^2 \hat y -4xy \hat z

Explanation:

Given the vector function

\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z

We can calculate the curl using the definition

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|

Thus for the exercise we will have

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

Working with the partial derivatives we get the curl

\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

6 0
3 years ago
The specific heat of a certain type of metal is 0.128 J/(g⋅∘C).0.128 J/(g⋅∘C). What is the final temperature if 305 J305 J of he
Makovka662 [10]

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

6 0
3 years ago
A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
3 years ago
Identify the arrows that show the movement of carbon from the biosphere to the atmosphere in the carbon cycle.
Illusion [34]

Answer:

All of the arrows pointing up that have a red box NOT the arrow pointing down with a red box. (if the blue squigglies on the water are arrows then they count too, the picture is not too clear)

Explanation:

8 0
3 years ago
Other questions:
  • How does charging by conduction occur?
    14·2 answers
  • The helicopter was deformed and destroyed in the _______ collision.
    15·2 answers
  • As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens th
    11·1 answer
  • The number of energy levels to which an electron can jump depends on the
    11·1 answer
  • B. A car moving at an initial speed vi applies its brakes and skids for some distance until coming to a complete stop. If the co
    9·1 answer
  • What is the meaning of the saying "the power of a lens is 1 dioptre "​
    11·1 answer
  • The speed of a wave with a frequency of 2 Hz (2/s), an amplitude of 3 m, and a wavelength of 10 m is
    11·1 answer
  • Is solar system a natural system or human-made system​
    13·2 answers
  • Between ball B and ball C, ball ____ has LESS potential energy because ___
    14·1 answer
  • A car is filled up with 20 gallons of gas. The car uses .25 gallons per minute. How much time will the car travel ?​
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!