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Tamiku [17]
3 years ago
15

What is the net force acting on the buggy? N The net force is pointing to the...

Physics
2 answers:
blagie [28]3 years ago
5 0

Answer:

390, right

Explanation:

antoniya [11.8K]3 years ago
4 0

Answer:

390,right

Explanation:

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Weigt A box sitting still on the ground by itself has a weight of 700N, what is the mass? (gravity's acceleration is 9.80 m/s2)​
Alex
  • Force=700N
  • Acceleration=9.8m/s^2

Using newtons law

\\ \rm\longmapsto F=ma

\\ \rm\longmapsto m=\dfrac{F}{a}

\\ \rm\longmapsto m=\dfrac{700}{9.8}

\\ \rm\longmapsto m=71.4kg

7 0
3 years ago
True or False:In a current, electrons will always flow from negative to positive.
DIA [1.3K]
It should be true: electrons low from negative toward positive to negative toward positive because opposite charges attract each other.

I hope this was correct
3 0
3 years ago
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Interactive LearningWare 4.1 reviews the approach taken in problems such as this one. A 1800-kg car is traveling with a speed of
Lubov Fominskaja [6]

Answer:

F= 4788 N

Explanation:

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

vf²=v₀²+2*a*d Formula (1)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s    

vf: final speed in m/s  

a: acceleration in m/s²

Data

d=36.9 m

v₀=14.0 m/s m/s    

vf= 0  

Calculating of the acceleration of the car

We replace dta in the formula (1)

vf²=v₀²+2*a*d

(0)²=(14)²+2*a*(36.9)

-(14)²= (73.8) *a

a= - (196) /  (73.8)

a= - 2.66 m/s²

Newton's second law of the car in direction  horizontal (x):

∑Fx = m*ax Formula (2)

∑F : algebraic sum of the forces in direction x-axis (N)

m : mass (kg)

a : acceleration  (m/s²)

Data

m=1800 Fkg

a= - 2.66 m/s²

Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :

We replace data in the formula (2)

-F= (1800 kg) * ( -2.66 m/s² )

F= 4788 N

6 0
3 years ago
HELP ME PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE A student uses a spring scale attached to a textbook to compa
aleksandrvk [35]

Answer:

1,860

Explanation:

4 0
3 years ago
An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45° relative to edge of the sidewalk. If it has
Marat540 [252]

Answer:14 s

Explanation:

Given

Velocity of ant is 0.1 m/s in a direction 45^{\circ}

if it has traveled 1 m perpendicular to the edge of the sidewalk

i.e. from diagram

\sin 45=\frac{1}{L}

L=\sqrt{2}

time=\frac{Distance}{speed}

t=\frac{\sqrt{2}}{0.1}

t=14.14 s

3 0
3 years ago
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