All categories

3 years ago

Please help on these questions!!!

Physics

1 answer:

tatiyna3 years ago

8 0

-- First question . . . first answer choice

-- Second question . . . second answer choice

-- Third question . . . third answer choice

You might be interested in

A dog runs at 35 m/s at 45 degrees N of E. What are its x and y components (all answers

pickupchik [31]

Answer:

its x and y component is 24.749m/s

Explanation:

Given

Speed of the dog = 35m/s

x component of the speed = xcos theta

y component of the speed = ycos theta

Given theta =45 degrees

x-component = 35cos45

x-component = 35(0.7071)

x-component = 24.749m/s

y-component = 35sin45

y-component = 35(0.7071)

y-component = 24.749m/s

Hence its x and y component is 24.749m/s

5 0

3 years ago

What do psychologists used to assist them in diagnosing abnormal behavior?

soldier1979 [14.2K]

Ok so it usually includes the evaluation of symptom and disorder severity, patterns of symptoms over time number, frequency, and duration of episodes, and the patient's strengths and weaknesses.

8 0

3 years ago

You push on a box and are unable to move it. What force opposes your push

Molodets [167]

<span> static friction :) your very welcome

</span>

7 0

3 years ago

An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part

guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

$E=\dfrac{kq}{r^2}$

k is electrostatic constant

if r = 2 m, electric field is given by :

$E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)$

If r = 1 m, electric field is given by :

$E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)$

Dividing equation (1) and (2) we get :

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1$

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0

3 years ago

A cube of copper 2.00cm on a side is suspended by a string. The cube is heated with a burner from 20.0°C to 90.0°C. The air surr

riadik2000 [5.3K]

Answer:

a. 2.856 x 10^-6 m³

b. 0.288 J

c. 971.88 J

d. 971.592 J

Explanation:

d) The change in internal energy of the cube.

Note: For Copper -

B(beta)= 5.1 × 10^-3 (C)^-1

Cp = 390J/Kg.K

p (Rho) = 8.90 × 10³Kg/m³

Detailed solution is attached below

5 0

3 years ago