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qwelly [4]
3 years ago
6

A force of 265.1 N acts on an object to produce an acceleration of 14.52 m/s^2. What is the mass of the object?

Physics
2 answers:
astra-53 [7]3 years ago
6 0

The answer is :


18.26


Hope I helped.

Lady bird [3.3K]3 years ago
4 0

Answer:

The mass of the object is 18.26kg

Explanation:

We know by Newton's second law we know that a force applied on an object is equal to its mass times the acceleration caused by the force:

F=ma

From this equation, solving for the mass:

m=\frac{F}{a}

Plugging in the values:

m=\frac{265.1N}{14.52m/s^2}=18.26kg

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Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie
elena-14-01-66 [18.8K]
The electric field generated by a point charge is given by:
E= k_e \frac{Q}{r^2}
where
k_e = 8.99 \cdot 10^9 Nm^2 C^{-2} is the Coulomb's constant
Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them. 

Let's calculate first the electric field generated by the positive charge at that point:
E_1=k_e  \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C
where the negative sign means its direction is toward the charge.

If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
E_{tot} =E_1 + E_2 = 7.19 \cdot 10^6 N/C+7.19 \cdot 10^6 N/C=1.44 \cdot 10^7 N/C
3 0
3 years ago
Suppose you wish to construct a motor that produces a maximum torque whose magnitude is 1.7 × 10-2 N·m. The coil of the motor ha
Len [333]

Answer:

86 turns

Explanation:

Parameters given:

Magnetic torque, τ = 1.7 * 10^(-2) Nm

Area of coil, A = 9 * 10^(-4) m²

Current in coil, I = 1.1 A

Magnetic field, B = 0.2 T

The magnetic toque is given mathematically as:

τ = N * I * A * B

Where N = number of turns

To find the number of turns, we make N subject of formula:

N = τ/(I * A * B)

Therefore:

N = (1.7 * 10^(-2)) / (1.1 * 9 * 10^(-4) * 0.2)

N = 85.85 = 86 turns (whole number)

The number of turns must be 86.

3 0
3 years ago
What part of the crane is used to lift and lower the load?
Flura [38]
I think the answer is b.boom
7 0
3 years ago
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Which evidence supports Wegener's hypothesis? Choose the three that
Tamiku [17]

I tried to look it up but it was saying that the super climate later broke apart in the sediments have been moving into their current positions ever since

8 0
3 years ago
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

\omega =\sqrt{\dfrac{K}{m}}

The maximum speed v given as

v=\omega A

A=Amplitude

v=\sqrt{\dfrac{K}{m}}\times A

0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

A=0.075 m

A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}

v= 0.21 m/s

5 0
3 years ago
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