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alukav5142 [94]

3 years ago

13

Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from t

he horizontal, and that the projectile is launched from the origin over a horizontal surface An ideal projectile is launched from level ground at a launch angle of 26° and an initial speed of 48 m/sec. How far away from the launch point does the projectile hit the ground

1 answer:

kirill115 [55]3 years ago

3 0

Answer:

d = 185.26 meters

Explanation:

It is given that,

Launching angle of the projectile, $\theta=26^{\circ}$

Initial speed of the projectile, u = 48 m/s

Let at distance d the projectile hits the ground from the launch point. It is equal to range of the projectile. Its formula is given by :

$d=\dfrac{u^2\ sin2\theta}{g}$

Substituting all the values in above formula. So, we get :

$d=\dfrac{(48)^2\ sin2(26)}{9.8}$

d = 185.26 meters

So, the distance between the launch point and the point where it hits is 185.26 meters. Hence, this is the required solution.

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A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

$K_{1} + W_{F} = K_{2}$

Where:

$W_{F}$ - Work done by the external force, measured in joules.

$K_{1}$ , $K_{2}$ - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

$W_{F} = K_{2} - K_{1}$

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$

Where:

$m$ - Mass of the object, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

$v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}$

$v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}$

$v_{1} \approx 25.060\,\frac{m}{s}$

Final velocity

$v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}$

$v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}$

$v_{2} \approx 33.121\,\frac{m}{s}$

Finally, if $m = 3.5\,kg$ , $v_{1} \approx 25.060\,\frac{m}{s}$ and $v_{2} \approx 33.121\,\frac{m}{s}$ , then the work done by the force is:

$W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]$

$W_{F} = 820.745\,J$

The work done by the force is 820.745 joules.

6 0

3 years ago

A bodybuilder deadlifts a 215 kg weight to a height of 0.90 m above the ground. If he deadlifts this weight 10 times in a span o

wariber [46]

A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)

<h3>What is power?</h3>

In physics, power (P) is the work (W) done over a period of time.

Step 1. Calculate the work done by the bodybuilder each time.

The bodybuilder lifts a 215 kg (m) weight to a height of 0.90 m (h). Being the gravity (g) of 9.81 m/s², we can calculate the work done in each lift using the following expression.

W = m × g × h = 215 kg × 9.81 m/s² × 0.90 m = 1.9 × 10³ N

Step 2. Calculate the work done by the bodybuilder over 10 times.

W = 10 × 1.9 × 10³ N = 1.9 × 10⁴ N

Step 3. Calculate the power exerted by the bodybuilder.

The bodybuilder does a work of 1.9 × 10⁴ N in a 45-s span.

P = 1.9 × 10⁴ N/45 s = 421 W

A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)

Learn more about power here: brainly.com/question/911620

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1 year ago

A calcium bromide compound is represented by which formula?

ipn [44]

Answer:

Calcium bromide is the name for compounds with the chemical formula CaBr2(H2O)x.

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3 years ago

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What is the slowest type of radiation?

tekilochka [14]

Velocity is the answer

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3 years ago

A block with mass m =6.9 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.24 m.

Burka [1]

U need to set up n solve the general eqn for simple harmonic motion:
x" = -(k/m)x

solution is x(t) = (x0)*cos(wt) + (v0/w)*sin(wt)
where w=sqrt(k/m), x0 is x-position at t=0 and v0 is vel at t=0
u already calculated f in Q.2 and w = 2*pi*f
x0 is 0 as it starts at eqm
v0 is given at 5.1

so u have x(t)

vel is given by x'(t) = (x0)*(-w)*sin(wt) + (v0/w)*w*cos(wt)

substitute t=0.32, x0=0, v0=5.1 n w in the above, u can solve for v at t=0.32.

5 0

3 years ago