All categories

3 years ago

What is the mass of an object moved with a force of 540 N at an acceleration of 7.74 m/s2?

Physics

2 answers:

goldenfox [79]3 years ago

8 0

Fnet=ma
Re arrange to m=fnet/a
M=540/7.74
M=69.76kg

Nataly_w [17]3 years ago

5 0

okay u need to use the equation m=Fnet/a. Fnet=540N a=7.74m/s^2 m=540N/7.74m/s^2=69.77kg

You might be interested in

Which of the following is a negatively charged particle that is found in "clouds" around the nucleus?

shutvik [7]

The answer is electron

Explanation: An atom consists of a small but massive nucleus that include protons (positive charge) and neutrons (neutral charge). around the nucleus is a cloud of rapidly moving electrons (negative charge)

3 0

3 years ago

Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take

KiRa [710]

Answer:

$\frac{V_{e}}{V_{h}}=0.428*10^{2}$

Explanation:

From conservation of energy states that

$K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}$

5 0

3 years ago

A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge

Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is $\lambda$ and the net charge per unit length is $2 \lambda$ .

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let $\lambda_i\ and\ \lambda_o$ are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

$\phi=\dfrac{q_{enclosed}}{\epsilon_o}$

$E.A=\dfrac{q_{enclosed}}{\epsilon_o}$

$0=\dfrac{\lambda_i+\lambda}{\epsilon_o}$

$\lambda_i=-\lambda$

For outer surface,

$\lambda_i+\lambda_o=2\lambda$

$-\lambda+\lambda_o=2\lambda$

$\lambda_o=3\lambda$

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

$E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}$

$E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}$

Hence, this is the required solution.

6 0

3 years ago

A loop of current-carrying wire has a magnetic dipole moment of 5. 0 10–4 am2. if the dipole moment makes an angle of 57° with a

Digiron [165]

The potential energy will be 1.46*10^-4J.

To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.

<h3>How to find the potential energy of the loop?</h3>

We have the expression for torque acting on a current loop in a uniform magnetic field as,

$\tau=MBsin\theta$

where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.

As we know that, the torque is equal to force times the perpendicular distance. Thus, it is equivalent to the work done. This work is stored as the potential energy in the loop.
Thus, the potential energy will be,

$\tau=W=U=MBsin\theta=5*10^{-4}*0.35*sin57=1.46*10^{-4}J$