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Cloud [144]
4 years ago
10

A 200-kg object has a 50-Newton rightward net force being applied to it. What is the magnitude of the rightward acceleration on

the object m/s^2?
Physics
1 answer:
katovenus [111]4 years ago
8 0
The formula for acceleration is force divided by mass.

Force = 50 N

Mass = 200 kg

Acceleration = ???

\frac{50}{200} = \frac{1}{4}

\frac{1}{4} = 0.25

~~

Acceleration = 0.25 m/s²

~~

Next time use, f = ma in a triangle,

F

M A

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey
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A lab assistant has been given data with units from an experiment to enter into the computer. Unfortunately, there are only numb
san4es73 [151]

Answer:

the answer ia A

Explanation:

3 0
3 years ago
A student sees a newspaper ad for an apartment that has 1330 square feet (ft^2) or floor space. How many square meters of area a
Natasha2012 [34]
<span>=1330 ft^2 x (1m^2/3.25 ft^2)
=124 m^2

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6 0
3 years ago
A car accelerates uniformly in a straight line with acceleration 10m/s 2 and travels 150m in a time interval of 5s. How far will
spayn [35]

Explanation:

Given:

Acceleration of car = 10 m/s

Distance travelled in 5 sec = 150 m

To find:

Distance travelled in the next 5 seconds

Concept:

There are 2 ways to app this kind of questions .

Either , we can find the total distance travelled in 10 seconds and then subtract 150 m from it.

Whereas , you can find out the final velocity at the end of the 5th seconds. Using equation of motion, then get the distance travelled in next 5 seconds.

Calculation:

v² = u² + 2as

=> ( u + at)² = u² + 2as

=> u² + 2uat + a²t² = u² + 2as

=> 2uat + (at)² = 2as

=> 2u (10)(5) + (10 × 5)² = 2 (10)(150)

=> 100u = 3000 - 2500

=> 100u = 500

=> u = 5 m/s

Distance travelled in 10 seconds :

s = ut + ½at²

=> s = (5 × 10) + ½(10)(10)²

=> s = 50 + 500

=> s = 550 m

Distance travelled in the 2nd half will be :

d = 550 - 150

=> d = 400 m

So final answer is :

6 0
3 years ago
A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of
SVEN [57.7K]

Answer:

Acceleration due to gravity is 20 m/sec^2

So option (E) will be correct answer

Explanation:

We have given length of the pendulum l = 2 m

Time period of the pendulum T = 2 sec

We have to find acceleration due to gravity g

We know that time period of pendulum is given by

T=2\pi \sqrt{\frac{l}{g}}

2=2\times 3.14 \sqrt{\frac{2}{g}}

0.3184= \sqrt{\frac{2}{g}}

Squaring both side

0.1014= {\frac{2}{g}}

g=19.71=20m/sec^2

So acceleration due to gravity is 20 m/sec^2

So option (E) will be correct answer.

8 0
3 years ago
Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag
nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β  ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β  =  0,5/0,7

tan β  = 0,7142    then   β = arctan 0,7142      ⇒   β = 35 ⁰

cos  β = 0,81

d = √ (0,5)²  +  (0,7)²       d1stance between charges

d = √0,25 + 0,49

d =  √0,74  m

d = 0,86  m

Now Foce between two charges  is:

F  = K* q₁*q₂/ d²      (1)

Where  K  = 9*10⁹ N*m²/C²

q₁  =  2,5* 10⁻⁹C

q₂  = 3,0*10⁻⁹C

d²  = 0,74 m²

Plugging these values in (1)

F  =  9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74       [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹   [N]

And  Fx  =  F*cos  β

Fx  =  91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹  [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹  [N]

F = 147,78*10⁻⁹ [N]

8 0
3 years ago
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