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3 years ago

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A dart leaves a toy dart gun with initial velocity of 7.76 m/s, regardless of the angle it is fired. What is the maximum horizon

tal distance that the dart can travel? Assume that the dart gun is fired at ground level.

1 answer:

In-s [12.5K]3 years ago

4 0

Vf=vi plus 2 ad
0=7.76 + 2(9.8)d
d=0.395m

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Your Lead Teaching Assistant is initially seated on the top of a hemispherical ice mound of radius R = 30 m. Approximate the ice

saw5 [17]

Answer:

<em>Height = 5.65 km</em>

Explanation:

$3\pi r^2$ is the circumference or we can say measures the boundary of hemisphere of friction-less ice that he is sitting on.

So, the height will be = 2 x 3.14 x (30)^2 = 5654.7 m = 5.65 km

7 0

3 years ago

Read 2 more answers

Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird

serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

db = 2dr
L + (L - x) = 2x
2L - x = 2x
2L = 3x
x = $\frac{2}{3}$ L

Insert it in x = $\frac{2}{3}$ L

$\frac{2}{3}$ (2.4km) = 1.6km

Now we use formula db = 2dr

db = 2L - x
db = 2(2.4km) - 1.6km
<u>db = 3.2km</u>

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

Vr = L/ Δt ⇒ Δt = $\frac{L}{Vr}$
$\frac{2.4km}{6.8km/hr}$
$\frac{6}{17}hr$

(Km cancel each other)

Vb = db/Δt ⇒ db = VbΔt
13.6km/hr $(\frac{6}{17}hr )$
<u>4.8km</u>

(hr cancel each other)

Hope it helps you :)

6 0

3 years ago

What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma

Sindrei [870]

Answer:

E = 1.50 × $10^{8}$ V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

$\mu _b = \frac{B}{2\mu _o}$ ...............1

and

energy density due to electric filed is

$\mu _e = \frac{\epsilon _o E^2}{2}$ ...............2

and here $\mu _b = \mu _ e$

so that

E = $\frac{B}{\sqrt{\mu _o \times \epsilon _o}}$ ...................3

put here value and we get

$E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}$

E = 3 × $10^{8}$ × 0.50

E = 1.50 × $10^{8}$ V/m

6 0

3 years ago

A 73-kg Norwegian olympian ski champion is going down a hill sloped at 39 ◦ . The coefficient of kinetic friction between the sk

bazaltina [42]

Answer:

Explanation:

net force on the skier = mg sin 39 - μ mg cos39

mg ( sin39 - μ cos39 )

= 73 x 9.8 ( .629 - .116)

= 367 N

impulse = net force x time = change in momentum .

= 367 x 5 = 1835 kg m /s

velocity of the skier after 5 s = 1835 / 73

= 25.13 m /s

b )

net force becomes zero

mg ( sin39 - μ cos39 ) = 0

μ = tan39

= .81

c )

net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s

so he will have speed of 25.13 m /s after 5 s .

5 0

3 years ago

A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time inte

musickatia [10]

Answer:

Number of electrons, $n=5.62\times 10^{21}$

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, $I=\dfrac{q}{t}$

$I\times t=n\times e$

$n=\dfrac{It}{e}$

e is the charge of an electron

$n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}$

$n=5.62\times 10^{21}$

So, the number of electrons pass through the resistor is $5.62\times 10^{21}$ . Hence, this is the required solution.

6 0

3 years ago