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tankabanditka [31]
3 years ago
13

Vocabulary Matching

Physics
1 answer:
Natasha_Volkova [10]3 years ago
5 0

Answer:

instruments

Explanation:

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The acceleration of a particle is given by ax(t) = -(2.00 m/s2) + (2.70 m/s3)t. (a) find the initial velocity v0x, such that the
Alexeev081 [22]
I’ll solve them right now just wait a few
3 0
3 years ago
A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of b
lakkis [162]

Answer:

T = 17.26 ^oC

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

Q_1 = Q_2

m_1s_1\Delta T_1 = m_2s_2\Delta T_2

so we will have

32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)

so we have

(45.8 - T) = 15.1(T - 15.4)

so we have

16.1 T = 277.87

T = 17.26 ^oC

6 0
3 years ago
A rocket has landed on planet x, which has half the radius of earth. An astronaut onboard the rocket weighs twice as much on pla
Nastasia [14]

Answer:

Option (c) u0

Explanation:

The escape velocity has a formula as:

V = √(2gR)

Where V is the escape velocity,

g is the acceleration due to gravity

R is the radius of the earth.

Now, from the question, we were told that the escape velocity for the rocket taking off from earth is u0 i.e

V(earth) = u0

V(earth) = √(2gR)

u0 = √(2gR) => For the earth

Now, let us calculate the escape velocity for the rocket taking off from planet x. This is illustrated below below:

g(planet x) = 2g (earth) => since the weight of the astronaut is twice as much on planet x as on earth

R(planet x) = 1/2 R(earth) => planet x has half the radius of earth

V(planet x) =?

Applying the formula V = √(2gR), the escape velocity on planet x is obtained as follow:

V(planet x) = √(2g(x) x R(x))

V(planet x) = √(2 x 2g x 1/2R)

V(planet x) = √(2 x g x R)

V(planet x) = √(2gR)

The expression obtained for the escape velocity on planet x i.e V(planet x) = √(2gR), is exactly the same as that obtained for the earth i.e V(earth) = √(2gR)

Therefore,

V(planet x) = V(earth) = √(2gR)

But from the question, V(earth) is u0

Therefore,

V(planet x) = V(earth) = √(2gR) = u0

So, the escape velocity on planet x is u0

4 0
4 years ago
A resistor with an unknown resistance is connected in parallel to a 13 Ω resistor. When both resistors are connected in parallel
larisa86 [58]

Answer:

R2 = 10.31Ω

Explanation:

For two resistors in parallel you have that the equivalent resistance is:

\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\      (1)

R1 =  13 Ω

R2 = ?

The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}            (2)

V: emf source voltage = 23 V

I: current = 4 A

You calculate the Req by using the equation (2):

R_{eq}=\frac{23V}{4A}=5.75\Omega

Now, you can calculate the unknown resistor R2 by using the equation (1):

\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega

hence, the resistance of the unknown resistor is 10.31Ω

8 0
3 years ago
Read 2 more answers
A ball moving at positive 3.0 m per s along a table rolls off a table and lands on the ground 2.0 m away. How high was the table
MA_775_DIABLO [31]

consider the motion along the horizontal direction :

v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s

X = horizontal displacement of the ball = 2.0 m

a = acceleration along the horizontal direction = 0 m/s²

t = time taken to land = ?

using the kinematics equation

X = v₀ t + (0.5) a t²

2.0 = 3.0 t + (0.5) (0) t²

t = 2/3


consider the motion of the ball along the vertical direction

v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s

Y = vertical displacement of the ball = height of the table = h

a = acceleration along the vertical direction = 9.8 m/s²

t = time taken to land = 2/3

using the kinematics equation

Y = v₀ t + (0.5) a t²

h = 0 t + (0.5) (9.8) (2/3)²

h = 2.2 m


C 2.2 m

3 0
3 years ago
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