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cluponka [151]
3 years ago
6

Which will have a larger momentum when moving at the same speed: a 2,000-kg truck or a 1,000-kg sedan

Physics
1 answer:
Alina [70]3 years ago
5 0

Answer:

2000 kg

Explanation:

Given that Which will have a larger momentum when moving at the same speed: a 2,000-kg truck or a 1,000-kg sedan

According to the definition of momentum, momentum is the product of mass and velocity.

That is,

Momentum = mass × velocity

Since velocity or speed is the same, then, the one of higher mass will have a greater momentum.

Therefore, the 2000 kg truck will have the greater momentum.

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The driver sees that the road is empty and accelerates at 1.0 m/s2 for 5.0 s. what can you determine about the truck's motion us
vekshin1

kinematic equation

v=u+at

v-u=at

v-u = 1x5

the driver will have increased speed by 5 m/s. actual speeds unknown

7 0
3 years ago
Which type of disease cannot be spread from one person to another
aniked [119]

I believe that would be D.

Cardiovascular disease generally refers to conditions that involve narrowed or blocked blood vessels that can lead to a heart attack, chest pain, or stroke. Which it has nothing to do with being infectious.

7 0
3 years ago
In general, how can you increase the rate of a chemical reaction?
lesantik [10]

Answer:

its 1

Explanation:

Several factors can increase the rate of a chemical reaction. In general, anything that increases the number of collisions between particles will increase the reaction rate, and anything that decreases the number of collisions between particles will decrease the chemical reaction rate.

Hope this helps? :))

3 0
3 years ago
Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 are approximations of the magnitude of the electric field of an electric dipole,
tamaranim1 [39]

Answer:

The ratio of E_{app} and E_{act} is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}

E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}

Second equation is

E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}

We need to calculate the ratio of E_{act} and E_{app}

Using formula

\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}

\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}

Put the value into the formula

\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}

\dfrac{E_{app}}{E_{act}}=0.9754

Hence, The ratio of E_{app} and E_{act} is 0.9754

8 0
3 years ago
An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force(emf) 9.0V and negl
mina [271]

Answer:

0.4 ohms.

Explanation:

From the circuit,

The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.

1/R' = 1/R1+1/R2

R' = (R1×R2)/(R1+R2)

R' = (2.4×1.2)/(2.4+1.2)

R' = 2.88/3.6

R' = 0.8 ohms.

Hence the current flowing through the circuit is

I = V'/R'................ Equation 1

Where V' = voltmeter reading

I = 6/0.8

I = 7.5 A

This is the same current that flows through the variable resistor.

Voltage drop across the variable resistor = 9-6 = 3 V

Therefore, the resistance of the variable resistor = 3/7.5

Resistance = 0.4 ohms.

7 0
3 years ago
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