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Sveta_85 [38]
3 years ago
14

11.76 cm² into two significant figures and include the appropriate units

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
8 0
You will need to turn this into scientific notation, so:

1.176 × 10^1 cm^2

To two significant figures means to two digits so since 7 is greater than 5, you will need to round up.

Final answer is:

1.2×10^1 cm^2
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A charge of 4.5x10^-5 C is placed in an electric field with a strength of 2.0x10^4 N/C. What is the electric force acting on the
Alex

Answer:

0.9 N

Explanation:

The electric force acting on a charge is given by:

F=qE

where

q is the magnitude of the charge

E is the strength of the electric field

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q=4.5\cdot 10^{-5}C is the charge

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Substituting into the equation, we find

F=(4.5\cdot 10^{-5}C)(2.0\cdot 10^4 N/C)=0.9 N

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two friends are talking to another person across the building the other person can hear them but not see them why is this?
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4 0
3 years ago
A constant force of 2.5 N to the right acts on a 4.5 kg mass for 0.90 s.
Alborosie

Answer:

(a) v_f=0.5\frac{m}{s}

(b) v_f=-11\frac{m}{s}

Explanation:

(a) Since a constant external force is applied to the body, it is under an uniformly accelerated motion. Using the following kinematic equation, we calculate the final velocity of the mass  if it is initially at rest(v_0=0):

v_f=v_0+at\\v_f=at(1)

According to Newton's second law:

F=ma\\a=\frac{F}{m}(2)

Replacing (2) in (1):

v_f=\frac{F}{m}t\\v_f=\frac{2.5N}{4.5kg}(0.9s)\\v_f=0.5\frac{m}{s}

(b) In this case we have v_0=-11.5\frac{m}{s}. So, we use the final velocity equation:

v_f=v_0+at\\v_f=v_0+\frac{F}{m}t\\v_f=-11.5\frac{m}{s}+\frac{2.5N}{4.5kg}(0.9s)\\v_f=-11\frac{m}{s}

8 0
3 years ago
Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 fe
nignag [31]

Answer:

a) x = 40 t , y = 39 t ,  z = 6 + 32 t - 16 t ²,   b)     x = 80 feet ,  y = 78 feet , the ball came into the field  

Explanation:

a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis

Since the cast is in the center of the field, let's place the coordinate system

          x₀ = 0

          y₀ = 0

          z₀ = 6 feet

x-axis (towards end zone,  GOAL zone)

         x = xo + v₀ₓ t

        x = 40 t

y-axis (field width)

        y = y₀ + v_{oy} t

        y = 39 t

z axis (vertical)

        z = z₀ + v_{oz} t - ½ g t²

        z = 6 + 32 t - ½ 32 t²

        z = 6 + 32 t - 16 t ²

b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive

          z = 6

          6 = 6 + 32 t - 16 t²

          (t - 2)t  = 0

           t=0 s

           t= 2 s

The ball position

          x = 40 2

          x = 80 feet

           

          y = 39 2

          y = 78 feet

the dimensions of the field from the coordinate system (center of the field) are

             x_total = 150 feet

             y _total = 80 feet

so we can see that the ball came into the field

6 0
2 years ago
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