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3 years ago

9

Hulk starts at -2 m, jumps to -15 m, and settles at 6 m. What is Hulk's displacement?

1 answer:

Lorico [155]3 years ago

6 0

Answer:

Sorry don't know the answer

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A 250. mL sample of gas at 1.00 atm and 20.0°C has the temperature increased to 40.0°C and the volume increased to 500. mL. What

ladessa [460]

Answer:

New pressure is 0.534 atm

Explanation:

Given:

Initial volume of the gas, V₁ = 250 mL

Initial pressure of the gas, P₁ = 1.00 atm

Initial temperature of the gas, T₁ = 20° C = 293 K

Final volume of the gas, V₂ = 500 mL

Final pressure of the gas = P₂

Final temperature of the gas, T₁ = 40° C = 313 K

now,

we know for a gas

PV = nRT

where,

n is the moles

R is the ideal gas constant

also, for a constant gas

we have

(P₁V₁/T₁) = (P₂V₂/T₂)

on substituting the values in the above equation, we get

(1.00 × 250)/293 = (P₂ × 500)/313

or

P₂ = 0.534 atm

Hence, the <u>new pressure is 0.534 atm</u>

5 0

3 years ago

Hulk starts at -2 m, jumps to -15 m, and settles at 6 m. What is Hulk's displacement?

Lorico [155]

Answer:

Sorry don't know the answer

6 0

3 years ago

A resistor is connected to a 36v power supply. An ammeter measures a current of 2.0 A going through it. Determine the resistance

m_a_m_a [10]

R = 18 ohms

Explanation:

Given:

V = 36 volts

I = 2.0 A

R = ?

Use Ohm's law to solve for the resistance:

V = IR

or

R = V/I

= (36 volts)/(2.0 A)

= 18 ohms

8 0

2 years ago

Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

stepan [7]

We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
$F=ma$
where $F=-50 N$ is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
$a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2$

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
$a= \frac{v_f^2-v_i^2}{2d}$
where
$v_f=0$ is the final speed of the sled
$v_i=5 m/s$ is the initial speed
$d$ is the distance covered

By rearranging the equation, we find d:
$d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m$

3 0

3 years ago

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

stellarik [79]

Answer:

0.243 m/s

Explanation:

From law of conservation of motion,

mu+m'u' = V(m+m')................. Equation 1

Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.

make V the subject of the equation

V = (mu+m'u')/(m+m')................. Equation 2

Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s

Substitute into equation 2

V = (260000×0.32+52500×(-0.14))/(260000+52500)

V = (83200-7350)/312500

V = 75850/312500

V = 0.243 m/s

8 0

3 years ago