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S_A_V [24]
3 years ago
5

Is my answer correct plz explain i don't understand it.

Physics
1 answer:
lora16 [44]3 years ago
4 0

Answer:

yes that is correct

Explanation:

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The slope of a line on a distance-time graph represents _____. distance time displacement speed
lisabon 2012 [21]

Answer:

speed

Explanation:

The slope of a  line on any distance-time graph represents the speed of the object.

Velocity  only comes in when there is speed of the object in a particular direction.

4 0
3 years ago
What is the least value of an ammeter <br>Is it 1. 0.1<br>2. 0.10<br>3. 0.2<br>4. 0.5​
givi [52]

The least value varies depends on ammeter range. In the given question, ammeter range is not mentioned. So, the least value of an ammeter is 0.1 or 0.5 (depends on ammeter range).

<u>Explanation:</u>

The least value of an ammeter is the measure of the smallest number which can be observed in ammeter. So the least value will be varying depending upon its range. If we consider the range of ammeter is 30 A and the scale readings are 10 numbers, then the least value will be 30/10 = 3 A per scale.

So the least value is determined as \frac{Range of ammeter}{No.of scales}

So among the given options 0.1 is most suitable for an ammeter with range of 3 A with 30 divisions.

3 0
3 years ago
If you ride your bike 20 miles and it takes you 120 minutes,what is your average speed
Alexus [3.1K]
120 minutes=2 hours
20/2= 10mph
5 0
3 years ago
A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse
Elanso [62]

Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg :  crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight  (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)   

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax    , ax=0

282 N - 309.58 N = 90*a  

a=  (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

v_{f} = 0.51 \frac{m}{s}

8 0
3 years ago
2 pts
const2013 [10]

Answer:

I think the answer 1

Explanation:

im probably wrong too i dont know

5 0
3 years ago
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