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3 years ago

First answer will be made brainliest

Physics

1 answer:

barxatty [35]3 years ago

3 0

Answer:

1.75 is the unit rate so yeah

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Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o

seraphim [82]

Answer:

$\theta = 15^o \: or\: 75^o$

Explanation:

As we know that the formula of range is given as

$R = \frac{v^2sin2\theta}{g}$

now we know that

maximum value of the range of the projectile is given as

$R_{max} = \frac{v^2}{g}$

now we need to find such angles for which the range is half the maximum value

so we will have

$\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}$

$sin(2\theta) = \frac{1}{2}$

$2\theta = 30 or 150$

$\theta = 15^o \: or\: 75^o$

7 0

3 years ago

A golf ball is rolling in the grass. What must happen to stop the ball from continuing to roll?

slamgirl [31]

The net force must be zero

This is in accordance to Newton's first law, which states that any object in motion will remain in motion and any object at rest will remain at rest unless acted upon by an unbalanced force. An unbalanced force is one where the net force is not zero. If no unbalanced force is applied to a moving object, it will keep moving forever. The reason that we do not observe this in our daily lives is due to friction acting as the unbalanced force.

4 0

2 years ago

Which subatomic particle has a negative charge?

ValentinkaMS [17]

Answer:

C. $\displaystyle electron$

Explanation:

0 charge → <em>Neutron</em>

1 charge → <em>Proton</em>

I am joyous to assist you anytime.

4 0

3 years ago

Why mole is called fundamental unit.

gladu [14]

Explanation:

because it doesn't depend upon other unit like kg meter and second

4 0

2 years ago

Find the change in velocity of a 369 g hockey puck subject to the force shown below.

Elena-2011 [213]

Answer:

The change in velocity is 15.83 [m/s]

Explanation:

Using the Newton's second law we have:

ΣF = m*a

The force in the graph is 185 N, therefore:

$185=0.369*a\\Where\\a=acceleration made it by the force [m/s^2]$

$a=501.35[m/s^2]$

Now using the following kinematic equation:

$V^{2}=Vi^{2} + 2*a*(x-xi) \\where\\V=final velocity [m/s]\\Vi= initial velocity [m/s] = 0 the hockey disk is in rest when receives the hit.\\ x = Final position [m] = 0.4 m\\xi = initial position [m] = 0.15m\\$

Now replacing the values:

$V^{2}=0 + 2*501.35*(0.4-0.15)\\ \\V= 15.83[m/s]$

3 0

3 years ago