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Anastaziya [24]

2 years ago

13

A 15kg object is travelling 10m/s and hit a stationary 10kg object and stuck together. What is the final velocity of the objects

?

1 answer:

qaws [65]2 years ago

7 0

Answer:

first object final velocity =2m/s

second = 12m/s

Explanation:

i hope this will help you,..,

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Which material would provide the

Bogdan [553]

Answer:

C) Plastic

Explanation:

Copper, silver, and gold all conduct electricity. Therefore, the material that would provide the most resistance for electricity would be plastic.

8 0

3 years ago

A constant magnetic field passes through a single rectangular loop whose dimensions are 0.42 m x 0.54 m. The magnetic field has

olga_2 [115]

Answer:

$0.29887\ \text{V}$

Explanation:

A = Area = $0.42\times0.54\ \text{m}^2$

B = Magnetic field = 1.7 T

$\theta$ = Angle that magnetic field makes with the plane of the loop = $71^{\circ}$

t = Time = 0.42 s

EMF is given by

$\varepsilon=A\cos\theta\dfrac{dB}{dt}\\ =\dfrac{0.42\times 0.54\times\cos 71^{\circ}\times 1.7}{0.42}\\ =0.29887\ \text{V}$

The magnitude of the average emf induced in the loop is $0.29887\ \text{V}$ .

8 0

2 years ago

A 77.0 kg firefighter climbs a flight of stairs 8.0 m high. how much work is required? 0 incorrect: your answer is incorrect. j

Alex73 [517]

<span>6160 joules

to lift 1 newton 1 metre requires 1 joule
there are 10 newtons in one kilo

so 77(kg) x 8 (metres) x 10 (newtons/kilo) = 6160 joules</span>

5 0

3 years ago

You have created a document that uses the term Frank L. Wright over 30 times. You want to change every occurrence of this term t

fgiga [73]

Find and Replace dialog box is probably the most useful.

3 0

3 years ago

A)If your torsion balance deflects to an angle of 10° when your two spheres are 40cm apart, what angle will it deflect to when t

svp [43]

Answer:

Explanation:

Given

for $\theta=10^{\circ}$

Sphere are $d=40\ cm$

when sphere $d_2=10\ cm$ apart suppose deflection is $\theta _2$

We know

$F=k_t\cdot \theta$

Where F=force between charged particle

$\theta =$ Deflection

$F=\frac{kQ_1Q_2}{r^2}=k_t\cdot \theta$

$\theta =\frac{k}{k_t}\times \frac{Q_1Q_2}{r^2}----1$

thus $\theta \propto \frac{1}{r^2}$

for $\theta _2$

$\frac{\theta _1}{\theta _2}=(\frac{r_2}{r_1})^2$

$\theta _2=16\times \theta _1$

$\theta _2=160^{\circ}$

(b)for $10^{\circ}$ deflection Potential $v_1=8\ kV$

Electric Potential is $V=\frac{kQ}{r}$

$Q=\frac{V\cdot r}{k}$

where V=voltage

k=constant

r=distance between charges

Put value of Q in equation 1

$\theta =\frac{k}{k_t}\times \frac{V^2r^2}{k^2}$

$\theta =\frac{V^2r^2}{k\cdot k_t}$

thus $\theta \propto V^2$

therefore

$\frac{\theta _1}{\theta _2}=(\frac{V_1}{V_2})^2$

$\frac{10}{\theta _2}=(\frac{8}{4})^2$

$\theta _2=\frac{10}{4}=2.5^{\circ}$

5 0

2 years ago