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3 years ago

| idk what to put this as lol this is related to Christianity |

Physics

2 answers:

Margarita [4]3 years ago

5 0

I believe the answer is <u>C</u>

I hope this helps! Have a nice day, make sure to take care of yourself. You're loved <3

Masteriza [31]3 years ago

4 0

C I think but i have to have 20 characters so

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An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a

Liono4ka [1.6K]

Answer:

$16.25^{\circ}$

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Horizontal range is given by

$R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}$

The angle at which the arrow is to be released is $16.25^{\circ}$ .

4 0

2 years ago

A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of 0.11 m and a mass of 0.01

AnnZ [28]

Answer:

Explanation:

The magnetic force acting horizontally will deflect the wire by angle φ from the vertical

Let T be the tension

T cosφ = mg

Tsinφ = Magnetic force

Tsinφ = BiL , where B is magnetic field , i is current and L is length of wire

Dividing

Tanφ = BiL / mg

= .055 x 29 x .11 / .010 x 9.8

= 1.79

φ = 61° .

Tension T = mg / cosφ

= .01 x 9.8 / cos61

= .2 N .

5 0

2 years ago

The angular speed of the hour hand of a clock, in rad/min, is:___________

lapo4ka [179]

The angular speed is defined as:

where

$T=12*60=720min$

$\omega=\frac{2\pi}{720}$

$\omega=4.4**10^{-3} rad/min$

4 0

3 years ago

A cheetah can go from the state of rest to running at 20m/s in just two seconds. What is the Cheetahs average acceleration

babymother [125]

acceleration = change in velocity/change in time

so...

a = 20 m/s / 2 seconds

a = 10

hope that helps :)

P.S. found this from Brainly User, sometimes all you have to do is search to find the answer.

7 0

2 years ago

A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck

Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

$M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing})$ where $M_{truck}$ is the mass of the truck, $V_{truck}$ is velocity of the truck, $V_{common}$ is the common velocity of moving and standing truck after collision and $M_{standing}$ is the mass of the standing truck

Making $V_{truck}$ the subject we obtain

$V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}$

Substituting $M_{truck}$ as 25000 Kg, $V_{common}$ as 22.3 m/s, $M_{standing}$ as 2000 Kg we obtain

$V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s$

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0

3 years ago