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BARSIC [14]
3 years ago
11

Please helppp!!

Physics
1 answer:
ira [324]3 years ago
6 0

Answer:

15*

Explanation:

Turned 15* cuz Θ=Θ'

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Which of the following examples illustrates static friction?
vivado [14]

Answer:

A box sits stationary  on a ramp

Explanation:

Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.

Static force of friction is calculated as follows:

F= μη

F is static force of friction.

μ is the coefficient of static friction.

η is the normal force.

6 0
4 years ago
Which of these substances will form a basic solution in water? A. Ca(OH)2 B. H3PO4 C. H2SO4 D. HCl
podryga [215]
Basic solutions are hydroxides therefore the answer is A ca(OH)2
3 0
3 years ago
A 50-kg block is at rest on a 15o slope. A force of 250 N is acting on the block up the slope parallel to it. If the block does
Iteru [2.4K]

The minimum value of the coefficient of static friction between the block and the slope is 0.53.

<h3>Minimum coefficient of static friction</h3>

Apply Newton's second law of motion;

F - μFs = 0

μFs = F

where;

  • μ is coefficient of static friction
  • Fs is frictional force
  • F is applied force

μ = F/Fs

μ = F/(mgcosθ)

μ = (250)/(50 x 9.8 x cos15)

μ = 0.53

Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.

Learn more about coefficient of friction here: brainly.com/question/20241845

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6 0
2 years ago
An engineering team has come to the stage in the engineering design
SIZIF [17.4K]

Answer: D

Explanation: read the answers above and compare to your test

6 0
3 years ago
To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 57.0° above the ho
cupoosta [38]

Answer:

xf = 5.68 × 10³ m  

yf = 8.57 × 10³ m  

Explanation:

given data

vi = 290 m/s

θ = 57.0°

t = 36.0 s

solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0                            yi = 0

xf = ?                            yf = ?

vxi =  vicosθ               vyi = visinθ  

ax = 0                          ay = −9.8 m/s

now we solve x motion: that is

xf = xi + vxi × t + 0.5 × ax × t²     ............1

simplfy it we get

xf = 0 + vicosθ × t + 0

put here value and we get

xf = 0 + (290 m/s) cos(57) (36.0 s)

xf = 5.68 × 10³ m  

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ²     ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s)  ²

yf = 8.57 × 10³ m  

5 0
3 years ago
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