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Mama L [17]
3 years ago
8

A force of 8,480 N is applied to a cart to accelerate it at a rate of 26.5 m/s2. What is the mass of the cart?

Physics
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:

<h2>320</h2>

Just use F =ma where m is mass and a is acc.

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Classify the following matter
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Answer:

Woah

Explanation:

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Blizzard [7]

Answer:

0.2 J

Explanation:

The pendulum forms a right triangle, with hypotenuse of 50 cm and base of 30 cm.  The height of this triangle can be found with Pythagorean theorem:

c² = a² + b²

(50 cm)² = a² + (30 cm)²

a = 40 cm

The height of the triangle is 40 cm.  The height of the pendulum when it is at the bottom is 50 cm.  So the end of the pendulum is lifted by 10 cm.  Assuming the mass is concentrated at the end of the pendulum, the potential energy is:

PE = mgh

PE = (0.200 kg) (9.8 N/kg) (0.10 m)

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6 0
4 years ago
now suppose that we have attached not just two springs in series, but N springs. Write an equation that expresses the effective
sammy [17]

Answer:

 k_{eq} = \frac{k}{N}

Explanation:

For this exercise let's use hooke's law

         F = - k x

where x is the displacement from the equilibrium position.

        x = - \frac{F}{k}

if we have several springs in series, the total displacement is the sum of the displacement for each spring, F the external force applied to the springs

       x_ {total} = ∑ x_i

we substitute

       x_ {total} =  ∑ -F / ki

       F / k_ {eq} =  -F  \sum \frac{1}{k_i}

      \frac{1}{k_{eq}} = \frac{1}{k_i} 1 / k_ {eq} =  ∑ 1 / k_i

if all the springs are the same

     k_i = k

     \frac{1}{k_{eq}} = \frac{1}{k} \sum 1 \\

     \frac{1}{k_{eq} } =  \frac{N}{k}

     k_{eq} = \frac{k}{N}

6 0
3 years ago
In a tug of war, when one team is pulling with a force of 85 N and the other 40 N, what is the net
olya-2409 [2.1K]
85 N - 40 N = 45 N
And depending on direction the greater force is being pulled towards
4 0
3 years ago
What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma
Sindrei [870]

Answer:

E = 1.50 × 10^{8} V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

\mu _b = \frac{B}{2\mu _o}   ...............1

and

energy density due to electric filed is

\mu _e = \frac{\epsilon _o E^2}{2}     ...............2

and here \mu _b = \mu _ e

so that

E = \frac{B}{\sqrt{\mu _o \times \epsilon _o}}      ...................3

put here value and we get

E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}  

E = 3 × 10^{8}  × 0.50

E = 1.50 × 10^{8} V/m

6 0
3 years ago
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