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3 years ago

Question 1

Physics

1 answer:

kakasveta [241]3 years ago

3 0

Answer:

W = 600 J

Explanation:

We have,

According to attached figure,

Height of the inclined plane is 60 m

Force acting on the block is 10 N

It is required to find the work must be done against gravity to move it to the top of the incline. The work done is given by :

W = mgh

$W=F\times h\\\\W=10\times 60\\\\W=600\ J$

So, the work done against the gravity is 600 J.

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The following diagram represents a cart with an initial velocity of 1.0 m/s sliding along a frictionless track from point A

ki77a [65]

Answer:

point b

Explanation:

7 0

3 years ago

During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla

Butoxors [25]

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of $u=-40\ m/s$

ball returned with speed of $v=40\ m/s$

average Force imparted by racquet on the ball is given by

$F_{racquet}=\frac{m(v-u)}{\Delta t}$

where $m=mass\ of\ ball$

$\Delta t=$ time of contact of ball with racquet

$F_{racquet}=\frac{m(40-(-40))}{\Delta t}$

$F_{racquet}=\frac{80m}{\Delta t}-----1$

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

$F_{player}=\frac{m(0-40)}{\Delta t}$

$F_{player}=\frac{-40m}{\Delta t}-----2$

From 1 and 2 we get

$F_{racquet}=-2F_{player}$

Hence the magnitude of Force by racquet is twice the Force by player

5 0

3 years ago

A school bus moves slower and slower. Using what you have learned about forces, explain why the bus moves slower and slower.

Levart [38]

Answer:

I don't exactly know what you learned but it could be because of more friction or the bus was running out of gas.

4 0

2 years ago

PLEASE The weight of a girl is 600 N and the area of her one foot is 50 cm². What will be the pressure exerted by her o

Rus_ich [418]

Answer:

pressure=force/area

p=600/50

p=12

so 12Nm^-1

4 0

3 years ago

An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equil

Nataly [62]

Answer:

The amount of ice added in gram is 32.77g

Explanation:

This problem bothers on the heat capacity of materials

Given data

Mass of water Mw= 200g

Temperature of water θw= 25°c

Temperature of ice θice= 0°c

Equilibrium Temperature θe= 12°c

Mass of ice Mi=???

The specific heat of ice Ci= 2090 J/(kg ∘C)

specific heat of water Cw = 4186 J/(kg ∘C)

latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

heat heat loss by water = heat gained by ice

N/B let us understand something, heat gained by ice is in two phases

Heat require to melt ice at 0°C to water at 0°C

And the heat required to take water from 0°C to equilibrium temperature

Hence

MwCwΔθ=MiLi +MiCiΔθ

Substituting our data we have

200*4186*(25-12)=Mi*3.3x10^5+

Mi*2090(12-0)

837200*13=Mi*3.3x10^5+Mi*2090

10883600=332090Mi

Mi=10883600/332090

Mi= 32.77g

4 0

3 years ago