All categories

3 years ago

I need help with this

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

6 0

The work done=700 joules

The mechanical advantage=8

Explanation

1.The force with which the player throws the ball F=20 newtons

The distance traveled by the ball D=35 m

we have to calculate the work done by the ball

W=F*D

=20*35=700 N/m=700 joules

2. the input force exerted on the lever =25 N

output force of the lever=200 N

we have to calculate the mechanical advantage which is the ratio of output force to input force

MA=output /input

=200/25=8

You might be interested in

Solve the quadratic by quadratic formula below<br><br> x squared-14x=-49

lubasha [3.4K]

X^2-14x+49=0

Factor
(x-7)(x-7)

Check
-7-7=-14
-7*-7=49

Use ZPP
x-7=0
x=7

Final answer: x=7

5 0

3 years ago

Read 2 more answers

Brad and Angelina can mow their yard together with two lawn mowers in 30 min. When brad works alone it takes him 45 min. How lon

Kisachek [45]

Answer:

Wouldn't it be 45 minutes as well?

Step-by-step explanation:

If you think about it in a real life scenario, if one person does the job and then another person does the same job at the same speed wouldn't it be the same time? But if its not i think it would be 15 minutes because 45-30 = 15

6 0

3 years ago

Drag each label to the correct location on the table . Triangle EFG . Rectangle PQRS . Triangle TUV . Rectangle ABCD . Your choi

Tanya [424]

Answer:

there should be a pic or a diagram

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

3(20-b)=36<br> I need help on this question!!

Daniel [21]

3(20 - b) = 36
60 - 3b = 36
-3b = 36 - 60
-3b = - 24
b = -24/-3
b = 8 <==

8 0

3 years ago

The diagram shows the sector of a circle with the centre O and radius 6cm.

cricket20 [7]

Answer:

Area of the shaded region = 1.92 cm²

Step-by-step explanation:

From the picture attached,

Area of the shaded region = Area of the sector OMN - Area of the triangle OMN

Area of sector OMN = $\frac{\theta}{360}(\pi r^{2})$

Here, θ = Central angle of the sector

r = radius of the sector

Area of sector OMN = $\frac{50}{360}(\pi )(6)^2$

= 15.708 square cm

Area of ΔOMN = 2(ΔOPN)

Area of ΔOPN = $\frac{1}{2}(OP)(PN)$

Area of ΔOMN = OP × PN

In ΔOPN,

sin(25°) = $\frac{PN}{ON}$

PN = ONsin(25°)

= 6sin(25°)

= 2.536 cm²

cos(25°) = $\frac{OP}{ON}$

OP = ONcos(25°)

OP = 6cos(25°)

OP = 5.438 cm

Area of ΔOMN = 2.536 × 5.438

= 13.791 cm²

Area of the shaded region = 15.708 - 13.791 = 1.917 cm²

≈ 1.92 cm²

3 0

3 years ago