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AleksandrR [38]

3 years ago

11

A 10kg object rests on a frictionless surface when it is struck by a 300N force. At what rate will it accelerate?

2 answers:

maxonik [38]3 years ago

5 0

Answer:

I believe the answer is C. 0.3m/s/s

Explanation:

hope this helps :) lemme know if I'm correct

krok68 [10]3 years ago

4 0

Answer:

Answer option B: 30m/s/s.

Explanation:

Should be the answer. I'm sorry if my answer is wrong not 100% sure. I tried my best.

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When you enter a toll road, your ticket is stamped 1:00 p.m. When you leave, after traveling 55 miles, your ticket is stamped 2:

Viefleur [7K]

27.5 because of you divide the 55miles with the time you get your velocity which is the speed.

4 0

3 years ago

Read 2 more answers

One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p

Naddik [55]

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

$P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}$

$kg\dfrac{m^2}{s^2}=Joule$

$P=\dfrac{kgm^2}{s^3}$

So, the answer is D. the proper replacement unit for one joule per second

8 0

3 years ago

Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. T

alex41 [277]

Answer:

30643 J

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

t = Time taken = 1 ns

c = Speed of light = $3\times 10^8\ m/s$

$E_0$ = Maximum electric field strength = $1.52\times 10^{11}\ V/m$

A = Area = $1\ mm^2$

Magnitude of magnetic field is given by

$B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T$

Intensity is given by

$I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2$

Power, intensity and time have the relation

$E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J$

The energy it delivers is 30643 J

4 0

3 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

3 years ago

A 50.0-g Super Ball traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A

Setler [38]

Explanation:

Average acceleration is change in velocity over time.

a = Δv / Δt

a = (22.0 m/s − (-25.0 m/s)) / 0.00350 s

a = 13,400 m/s²

7 0

3 years ago