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Juliette [100K]
3 years ago
10

In an experiment performed in a space station, a force of 60n causes an object to have an acceleration equal to 4m/s s .what is

the objects mass?
Physics
1 answer:
Troyanec [42]3 years ago
5 0
We Know, F = m*a
Here, F = 60N
a = 4 m/s²

Substitute their values in the equation,
60 = m*4
m = 60/4
m = 15

So, your final answer & the mass of the object would be 15 Kg

Hope this helps!
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It is vital to keep in mind that breaches are entirely concerned with data. No matter what physical damage a device incurs, data
Ierofanga [76]

Answer:

TRUE

Explanation:

We currently live in the digital age, where almost everything is digitized, including strategic information. Thus, each individual and especially corporations that have sensitive data must use protection mechanisms against cyber attacks. One of the measures most recommended by experts is encryption, which consists of a set of rules that aims to encode data information so that only the sender and the receiver can decipher it.

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A typical cell has a membrane potential of -70 , meaning that the potential inside the cell is 70 less than the potential outsid
Oxana [17]

Answer:

energy is stored is 2.2 × 10⁻¹³ J

Explanation:

The capacitance  of the cell is given with the expression

C = (KE₀A) / d

k is the dielectric constant, A is the area of the cell, d is the thickness of the cell.

Now given that; the diameter is 50,

Area A = 4πR²

A = 4π × ( 25 × 10⁻⁶ m)²

A = 7850×10⁻¹² m²

our capacitance C = (KE₀A) / d

C = [9 ( 8.85 × 10⁻¹² C²/N.m²  × 7850×10⁻¹² m² )] / 7×10⁻⁹ m

C = 8.93 × 10⁻¹¹ F

Now Energy stored

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E = 2.2 × 10⁻¹³ J

Therefore energy is stored is 2.2 × 10⁻¹³ J

8 0
3 years ago
An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

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\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

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a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

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PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
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What is the equivalent resistance of a circuit that contains three 10.0 Ω
yanalaym [24]

Answer:

O D.30.0 Ω

Explanation:

this <em>is </em><em>the </em><em>correct </em><em>answer</em><em>!</em>

8 0
2 years ago
Read 2 more answers
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