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4 years ago

11

Zachary, a lab instructor, is writing a safety contract that his chemistry class will use for an experiment. He mentions that st

udents should not eat or drink in the lab. Which other item should also be included?

2 answers:

vlada-n [284]4 years ago

7 0

Answer:

do not chew gum in the lab

Explanation:

got it right on edgenuity :)

Cloud [144]4 years ago

3 0

Students should be sure to put on their goggles during chemical test.

Students should be sure to wear their hand gloves during chemical test.

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An artillery shell is launched on a flat, horizontal field at an angle of α = 31.7° with respect to the horizontal and with an i

kompoz [17]

Answer:193.90 m/s

Explanation:

Given

launch angle $\theta =31.7^{\circ}$

launch velocity $v_0=202 m/s$

Horizontal velocity of the shell after $t=18.96 s$

time of flight of Projectile $T=\frac{2u\sin \theta }{g}$

$T={\frac2\times 202\sin 31.7}{9.8}$

$T=21.66 s$

i.e. projectile is declining as $t>\frac{T}{2}$

but horizontal component of velocity will remain same as there is no opposing force in horizontal direction

Horizontal component of velocity is

$u_x=v_0\cos \theta =202\cdot \cos 31.7=193.90 m/s$

8 0

3 years ago

Read 2 more answers

A lightbulb has a resistance of 195 Ω and carries a current of 0.62 A.

ki77a [65]

Answer:

75

Explanation:

Power is current times voltage:

P = IV

Voltage is current times resistance:

V = IR

Therefore:

P = I²R

Given I = 0.62 A and R = 195 Ω:

P = (0.62 A)² (195 Ω)

P ≈ 75 W

8 0

3 years ago

A man walks 30m to the west, then 5m to the east in 45 seconds. What was his displacement?

AleksandrR [38]

The displacement is the distance from the beginning point to the ending point. The time was just to throw you off. If he walks 30m due west, and then 5m due east, his displacement would be 30-5=25m to the west.

6 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

3 years ago

Which of these does NOT result in the acceleration of an object?

Zanzabum

I guess the correct answer is the first one.

6 0

3 years ago

Read 2 more answers