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AnnZ [28]
3 years ago
11

Help please (about projectile motion at an angle)

Physics
1 answer:
PolarNik [594]3 years ago
7 0

Answer:

Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Explanation:

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A 3.0kg mass tied to a string
dem82 [27]

Answer:

\boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN}

Given:

Mass (m) = 3.0 kg

Uniform speed (v) = 20 m/s

Length of string (r) = 40 cm = 0.4 m

To Find:

Tension in the string (T)

Explanation:

Tension (T) is the string will be equal to centripetal force (\sf F_c).

\boxed{ \bold{ T = F_c  =  \frac{m {v}^{2} }{r} }}

Substituting value of m, v & r in the equation:

\sf \implies T =  \frac{3 \times  {20}^{2} }{0.4}  \\  \\  \sf \implies T = \frac{3 \times 400}{0.4}  \\  \\  \sf \implies T =3 \times 1000 \\  \\  \sf \implies T =3000 \: N \\  \\ \sf \implies T =3 \: kN

\therefore

Tension in the string (T) = 3 kN

5 0
3 years ago
An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit
lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is 1.127\times 10^7m/s

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

m_1v_1=m_2v_2

where,

m_1\text{ and }v_1 = Initial mass and velocity

m_2\text{ and }v_2 = Final mass and velocity

We are given:

m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s

Putting values in above equation, we get:

238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s

Hence, the velocity of released alpha particle is 1.127\times 10^7m/s

4 0
2 years ago
There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2
tangare [24]

Answer:

Explanation:

Given

speed of Electron u=2\times 10^7\ m/s

final speed of Electron v=4\times 10^7\ m/s

distance traveled d=1.2\ cm

using equation of motion

v^2-u^2=2as

where v=Final velocity

u=initial velocity

a=acceleration

s=displacement

(4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}

a=5\times 10^{16}\ m/s^2

acceleration is given by a=\frac{qE}{m}

where q=charge of electron

m=mass of electron

E=electric Field strength

5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}

E=248.3\ kN/C                

5 0
3 years ago
A balloon is expanded to the same volume as that of a human head. Do an order-of-magnitude estimate of the volume of this balloo
cestrela7 [59]

Answer:

Volume of balloon =  1000 cm^3

Explanation:

 The head of a normal person can be assumed as a sphere with radius 10 cm.

 Volume of sphere =\frac{4}{3} \pi r^3, where r is the radius.

 We have approximate radius = 10 cm.

  Approximate volume of head =\frac{4}{3} \pi r^3=\frac{4}{3} *\pi* 10^3=4188cm^3

 In the given options the closest value to the approximate volume is 1000 cm^3.

 So, volume of head = Volume of balloon =  1000 cm^3

4 0
2 years ago
Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters
amid [387]

Answer:

Part a)

t = 1.65 s

Part b)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

v = 27.3 m/s

direction of velocity is given as

[tex]\theta = 26.35 degree

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

a = \frac{3}{4}(9.81)

a = 7.36 m/s^2

now the vertical displacement covered by the canister is given as

y = 10 m

now by kinematics we have

y = \frac{1}{2}gt^2

10 = \frac{1}{2}(7.36)t^2

t = 1.65 s

Part b)

Horizontal speed of the canister is given as

v_x = 24.5 m/s

now the distance moved by it

x = v_x t

x = 24.5 (1.65)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

v_x = 24.5 m/s

final velocity in Y direction

v_y = v_i + at

v_y = 0 + (7.36)(1.65)

v_y = 12.14 m/s

now magnitude of velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{24.5^2 + 12.14^2}

v = 27.3 m/s

direction of velocity is given as

\theta = tan^{-1}\frac{v_y}{v_x}

\theta = tan^{-1}\frac{12.14}{24.5}

[tex]\theta = 26.35 degree

6 0
2 years ago
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