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asambeis [7]
2 years ago
14

A crate is pulled with a force of 165 N at an angle 30 ° northwest. What is the resultant horizontal force on the crate?

Physics
1 answer:
Ad libitum [116K]2 years ago
5 0

Answer:

Resultant horizontal force = 143 N

Explanation:

Since the a gle is 30° northwest, then it means the resultant force will be horizontal and as such;

Resultant horizontal force = 165 * cos 30

Resultant horizontal force = 142.89

Approximating to a whole number gives;

Resultant horizontal force = 143 N

You might be interested in
A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.
Kobotan [32]

Explanation:

The given data is as follows.

           F = 5.75 \times 10^{-16} N

          q = 1.6 \times 10^{-19} C

          v = 385 m/s

       sin (63.9^{o}) = 0.876

Now, we will calculate the magnitude of magnetic field as follows.

              B = \frac{F}{qv sin (\theta)}

                  = \frac{5.75 \times 10^{-16} N}{1.6 \times 10^{-19} C \times 385 m/s \times 0.876}

                  = 0.01065 \times 10^{3} T

                  = 10.65 T

Thus, we can conclude that magnitude of the magnetic field is 10.65 T.

4 0
3 years ago
g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. Fo
mariarad [96]

Answer:

A) ≥ 325Kpa

B) ( 265 < Pe < 325 ) Kpa

C) (94 < Pe < 265 )Kpa

D)  Pe < 94 Kpa

Explanation:

Given data :

A large Tank : Pressures are at 400kPa and 450 K

Throat area = 4cm^2 ,  exit area = 5cm^2

<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>

The range of flow of back pressures that will make the flow entirely subsonic

will be ≥ 325Kpa

attached below is the detailed solution

<u>B) Have a shock wave</u>

The range of back pressures for there to be shock wave inside the nozzle

= ( 265 < Pe < 325 ) Kpa

attached below is a detailed solution

C) Have oblique shocks outside the exit

= (94 < Pe < 265 )Kpa

D) Have supersonic expansion waves outside the exit

= Pe < 94 Kpa

7 0
3 years ago
2×3.14√(1.0m/(9.8〖ms〗^(-1) )=)
il63 [147K]

This is the period in a simple harmonic motion which is 2 seconds in this question.

<h3>What is Period ?</h3>

The period of an oscillatory object can be defined as the total time taken  by a vibrating body to make one complete revolution about a reference point.

We are given the below question

2×3.14√(1.0m/(9.8〖ms〗^(2) )= T

This question can as well be expressed as

2π√(L/g) which is equal to period T.

In a nut shell, Period T = 2×3.14√(1.0m/9.8)

T = 6.28√0.102

T = 6.28 × 0.32

T = 2.006 s

Therefore, the period T of the oscillation is 2 seconds approximately.

Learn more about Period here: brainly.com/question/12588483

#SPJ1

8 0
1 year ago
I NEED PHYSICS HELP ASAP! DO NOT GUESS! I WILL GIE BRAINLEIST IF RIGHT! THANKS! :D
ale4655 [162]
<span>The repelling of the support magnet decreases friction. is the answer you're looking for .  :)

hope i helped - beanz</span>
6 0
3 years ago
Read 2 more answers
Who think's im pretty <br> the pic is a lil hard to see
damaskus [11]

Answer:

which pic...? there is no picture attached to your question

6 0
3 years ago
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