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Minchanka [31]
2 years ago
7

Is it true or false When objects collide, some momentum is lost.

Physics
2 answers:
never [62]2 years ago
5 0

Answer:

True.

Explanation:

BabaBlast [244]2 years ago
5 0
This is true since they have something stopping it
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An object has an average acceleration of +6.18 m/s2 for 0.266 s . At the end of this time the object's velocity is +9.90 m/s .
Archy [21]

At the start of the 0.266 s, the object's speed was 8.26 m/s.

The question can only be talking about speed, not velocity.

4 0
3 years ago
A can of soup is begin heated in a sauce pan. as the soup is heated, the warm fluid becomes less dense and rises, while the cool
Alexxandr [17]
It is c convection because convection is when something is hot therefore less dense material to rise and cooler material to sink under the influence of gravity
8 0
3 years ago
I AM........ INEVITABLE
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Answer:

that's nice very nice super duper nicer

5 0
3 years ago
Thermal conductivity of a material is given as 129Btuft–¹ h–¹°F–¹.Calculate this thermal conductivity in Jm–¹s–¹°C–¹(Given: 1Btu
ss7ja [257]

Answer:

223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

$1^\circ F = \frac{5}{9}^\circ C$

We know that 1 h = 3600 s

So the thermal conductivity of the material in $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$ is  :

Thermal conductivity :

$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$

$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$

=  223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

3 0
3 years ago
A car of 900 kg mass is moving at the velocity of 60 km/hr. It is brought into rest at 50 meter distance by applying a brake. No
kozerog [31]

Answer: -2502N

Explanation:

(V_2)^2=(V_1)^2+2ad

where;

V_2 = final velocity = 0

V_1 = initial velocity = 60 km/h = 16.67 m/s

a = acceleration

d = distance

First all of, because acceleration is given in m/s and not km/h, you need to convert 60km/h to m/s. Our conversion factors here are 1km = 1000m and 1h = 3600s

60km/h(\frac{1000m}{1km} )(\frac{1h}{3600s} )=16.67m/s

Solve for a;

(V_2)^2=(V_1)^2+2ad

Begin by subtracting (V_1)^2

(V_2)^2-(V_1)^2=2ad

Divide by 2d

\frac{(V_2)^2-(V_1)^2}{2d} =a

Now plug in your values:

a=\frac{(0)^2-(16.67 m/s)^2}{2(50m)}

a=\frac{0-277.89m^2/s^2}{100m}

a=-2.78m/s

If you're wondering why I calculated acceleration first is because in order to find force, we need 2 things: mass and acceleration.

F=ma

m = mass = 900kg

a = acceleration = -2.78m/s

F=(900kg)(-2.78m/s)\\F=-2502N

It's negative because the force has to be applied in the opposite direction that the car is moving.

8 0
3 years ago
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