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1 year ago

What is the mass of an object that requires 100N (kg-m/s2) of force in order to accelerate it at 10m/s2 (Please use G-R-E-S-A)

Physics

1 answer:

anastassius [24]1 year ago

3 0

Answer:

10kg

Explanation:

(I'm not super familiar with the GRESA method so apologies for any inaccuracies)

Given: We are given values for Force: 100N, and Acceleration: 10m/s2.

Required: We are trying to find Mass (m)

Equation: The best equation to use to solve this problem is F=ma,

Force = Mass x Acceleration. We can rearrange this for mass: m = F/a.

Solution: By substituting in the values we have: $m = \frac{100}{10}$

Answer: Mass = 10kg

Hope this helped!

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A straight wire segment 2 m long makes an angle of 30degrees with a uniform magnetic field of 0.37 T. Find the magnitude of the

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Answer : 0.814 newton

Explanation:

force (magnetic) acting on the wire is given by

F= ? , I=2.2amp , B = 0.37 T

F = B i l sin (theta) = 0.37 x 2.2 x 2x 0.5 = 0.814N

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3 years ago

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An object with a mass of 0.5 kilometre start from rest and achieves a maximum speed of 20 metre per second in 0.01 second, what

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2 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

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3 years ago

What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an

astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution= $T_b=101^{\circ}$ C

Boiling point water=100 degree Celsius

$K_f=1.86K/m$

$K_b=0.512 K/m$

$\Delta T_b=T-T_0$

Where $T$ =Boiling point of solution

$T_0=$ Boiling point of pure solvent

$\Delta T_b=101-100=1^{\circ}$ C

$\Delta T_b=k_bm$

Using the formula

$1=0.512\times m$

Molality, $m=\frac{1}{0.512}$ m

$\Delta T_f=k_fm$

Using the formula

$\Delta T_f=\frac{1}{0.512}\times 1.86$

$\Delta T_f=3.63 C$

We know that

$\Delta T_f=T_0-T_1$

Where $T_0$ =Freezing point of solvent

$T_1=$ Freezing point of solution

Using the formula

$3.63=0-T_1$

Freezing point of water=0 degree Celsius

$T_1=0-3.63=-3.63 C$

Hence, the freezing point of solution=-3.63 degree Celsius

7 0

3 years ago

A ball is thrown vertically upward with a speed of 1.86

Inessa05 [86]

Answer:

t = 1.09 s.

Explanation:

This is a one-dimensional kinematics question, so the equations of kinematics will be sufficient to solve the question.

$y-y_0 = v_0t + \frac{1}{2}at^2\\0 - 3.82 = 1.86t +\frac{1}{2}(-9.8)t^2\\-3.82 = 1.86t - \frac{1}{2}9.8t^2\\4.9t^2 - 1.86t - 3.82 = 0$

This quadratic equation can be solved using determinant.

$\Delta = b^2 - 4ac\\t_{1,2} = \frac{-b \pm \sqrt{\Delta} }{2a}\\t_1 = 1.09~s\\t_2 = -0.71~s$

Of course, we will choose the positive time.

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3 years ago