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2 years ago

Please help me with my quiz

Physics

1 answer:

Oduvanchick [21]2 years ago

8 0

Answer:

matter: A

data: E

variable: C

Controlled: D

Physical: B

Explanation:

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ANSWER SOOONN PLZ!!

Ket [755]

c it is both a and b

3 0

2 years ago

A baseball is thrown horizontally at a rate of 40m/s toward a home plate 18.4 m away. How far below the launch height is the bal

sammy [17]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g where

R is the distance moves in horizontal direction = 18.4m

H is the height

U is the velocity of the baseball = 40m/s

g is the acceleration due to gravity = 9.8m/s²

Substitute the given parameters into the formula and calculate H as shown;

18.4 = 40√2H/9.8

18.4/40 = √2H/9.8

0.46 = √2H/9.8

square both sides;

(0.46)² = (√2H/9.8)²

0.2116 = 2H/9.8

2H = 9.8*0.2116

2H = 2.07368

H = 2.07368/2

H = 1.03684m

Hence the ball is 1.03684m below the launch height when it reached home plate.

8 0

2 years ago

A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe

liubo4ka [24]

Answer:

ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

$I_1=\dfrac{M+m}{2}r^2$

$I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2$

$I_1=0.75\ kg.m^2$

Final mass moment of inertia

$I_2=\dfrac{M}{2}r^2$

$I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2$

$I_2=0.468\ kg.m^2$

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

$\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s$

ω₂=1.20

7 0

2 years ago

Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep

Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = $\frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}$

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 - 0:00 = 5

D'(5) = -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 - 0:00 = 6

D'(5) = -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

D'(5) = -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 - 0:00 = 12

D'(5) = -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0

2 years ago

Que es una ondaaaa?? no me deja escribiiiirrrrrr

UkoKoshka [18]

Answer:

yo queiro escribir mi hermano

7 0

3 years ago