How is resonance used in musical instruments?

When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th

Fofino [41]

Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

3 0

2 years ago

Can anyone solve these for my by using unit vectors? Can you also please show your work

Oxana [17]

4. The Coyote has an initial position vector of $\vec r_0=(15.5\,\mathrm m)\,\vec\jmath$ .

4a. The Coyote has an initial velocity vector of $\vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath$ . His position at time $t$ is given by the vector

$\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2$

where $\vec a$ is the Coyote's acceleration vector at time $t$ . He experiences acceleration only in the downward direction because of gravity, and in particular $\vec a=-g\,\vec\jmath$ where $g=9.80\,\frac{\mathrm m}{\mathrm s^2}$ . Splitting up the position vector into components, we have $\vec r=r_x\,\vec\imath+r_y\,\vec\jmath$ with

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$r_y=15.5\,\mathrm m-\dfrac g2t^2$

The Coyote hits the ground when $r_y=0$ :

$15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s$

4b. Here we evaluate $r_x$ at the time found in (4a).

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m$

5. The shell has initial position vector $\vec r_0=(1.52\,\mathrm m)\,\vec\jmath$ , and we're told that after some time the bullet (now separated from the shell) has a position of $\vec r=(3500\,\mathrm m)\,\vec\imath$ .

5a. The vertical component of the shell's position vector is

$r_y=1.52\,\mathrm m-\dfrac g2t^2$

We find the shell hits the ground at

$1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s$

5b. The horizontal component of the bullet's position vector is

$r_x=v_0t$

where $v_0$ is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for $v_0$ :

$3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}$

5 0

3 years ago

A bowling ball is rolling in a straight line toward the pins at a speed of 2.2 m/s. Its momentum is 4.75 kg*m/s. What is the

Thepotemich [5.8K]

Answer: m= 2.16 kg

Explanation: Momentum is expressed in the following formula:

p = mv

Derive to find m:

m = p / v

= 4.75 kg.m/s / 2.2 m/s

= 2.16 kg

Cancel out m/s and the remaining unit is in kg.

3 0

3 years ago

Read 2 more answers

A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang

olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: $s_c = \omega_ct = 0.233t$