can anyone help find the length where the monkey is? When the horizontal rope support a tension of 110N. The mass oof monkey is 10kg and climbs up the 120N uniform ladder
1. can anyone help find the length where the monkey is? When the horizontal rope support a tension of 110N. The mass oof monkey is 10kg and climbs up the 120N uniform ladder
a.0.219 L
semoga membantu
2. Aleron measures the length of three ropes. Rope A is 1/2 m long, Rope B is 5/8 m long and Rope C is 3/8 m long. What is the total length of three ropes?The total length of three ropes is _______ m. * Jawaban Anda
Jawab:
1 1/2 m
Penjelasan dengan langkah-langkah:
Rope A = 1/2 m x4 = 4/8 m
Rope B = 5/8 m
Rope C = 3/8 m
Total = 4/8 m + 5/8 m + 3/8 m = 9/8 m + 3/8 m
= 12/8 m
= 1 4/8 m
= 1 1/2 m
Jawaban:
1 1/2
Penjelasan dengan langkah-langkah:
semoga membantu......
3. Diana measures the length of three ropes. Rope A is 1/2 m long.Rope B is 5/8m long and Rope C is 3/8 m long.What is the total length of the three ropes?
Jawaban:
One Meter and half long
4. A kitten and a basket has a mass of 3 kg. A hanster and a similar basket has a mass of 1.4 kg. The mass of the kitten is 5 times the mass of the hamster. (a) what is the mass of the hamster? (b) what is the mass of the kitten?
hamster = x
kitten =y
basket=z
y=5x
x + z = 1,4
y+ z = 3. (+)
x-y = -1,6
x-5x=-1,6
6x = 1,6. x =0,267
y = 5x=5×0,267=1,335
5. A metal bar of mass 3.6kg is cut into two pieces in the ratio 3:5. The length of the shorter piece is 45cm. Finda)the length of the longer piece. b)the length of the original metal. bar c)the mass per unit length of the bar in kg/md)the mass of the shorter piece.
The Ratio of length is 3 : 5
The length of the shorter piece = 45 cm
a) The length of the longer piece (x = ?)
[tex] \frac{3}{5}= \frac{45}{x} \\ \\ 3x = 45 \times 5 \\3x=225 \\ \\ x= \frac{225}{3} \\ \\ x= 75\ cm [/tex]
b) The length of the original is
= 45 cm + 75 cm
= 120 cm or 1,2 m
c) The mass per meter is
= 3,6 kg / 1,2 m
= 3 kg/m
d) The mass of the shorter piece
= 3 kg/m × 0,45 m
= 1,35 kg
6. ection C. (30 marks) Kathy had a rope of 60.5 m. She cut the rope into two pieces. If the length of one piece was 6.3 m shorter than the other one. What was the length of the longer piece of rope?
Jawaban:
= 60.5 - 6.3
= 54.2 m
= 54.2 / 2
= 27.1 m
= 27.1 + 6.3
= 33.4m
7. Rope A is 1/6 meter shorter than Rope C. Rope B is 1/5 meter longer than Rope C. The total length of the three ropes is 2-18/15 meters. a. How long is rope A ? b. How long is rope C ?
Jawaban:
[tex]a = c - \frac{1}{6} \\ b = c + \frac{1}{5} \\ \\ a + b + c = \frac{48}{15} \\ c - \frac{1}{6} + c + \frac{1}{5} + c = \frac{48}{15} \\ c + c + c = \frac{48}{15} + \frac{1}{6} - \frac{1}{5} \\ 3c = \frac{96 + 5 - 6}{30} = \frac{95}{30} \\ c = \frac{95}{30} \times \frac{1}{3} = \frac{95}{90} = \frac{19}{18} meters \\ \\ a). \: a = c - \frac{1}{6} \\ a = \frac{19}{18} - \frac{1}{6} = \frac{19 - 3}{18} = \frac{16}{18} = \frac{8}{9} meters \\ \\ b). \: c = \frac{18}{19} meters[/tex]
8. a kitten and a basket has a mass of 3 kg . A hamster and a similar basket has a mass of 1.4kg . the mass of the kitten is 5 times the mass of the hamster .a. what is the mass of the hamster ? , b. what is the mass of the kitten ?
kitten = x ,hamster = y and baskest = z
x+z = 3 kg
y+z = 1,4 kg
z = x- y = 3 kg - 1,4 kg = 1,6 kg
x+y = 4,4 kg - 1,6 kg = 2, 8 kg
x = 5y
5y +y = 2,8 kg
6y = 2,8 kg
y = 2,8 kg /6
y = 0,47 kg
x = 5y
= 5 * 0,47 kg
= 2,35 kg
a. mass off hamster = 0,47 kg
b. mas off kitten = 2,35 kg
bonus...mass of basket = 1,6 kg
thank's
9. Rope A is 1/6 meter shorter than Rope C. Rope B is 1/5 meter longer than Rope C The total length of the three ropes is 2 18/25
Penjelasan dengan langkah-langkah:
A = C - 1/6
B = C + 1/5
A + B + C = 2 18/15 = 48/5
A - C = -1/6
A + B + C = 48/15
-------------------------- +
2A + B = -1/6 + 48/15 = -5/30 + 96/30 = 91/30
4A + 2B = 2 × 91/30 = 91/15
B - C = 1/5
A + B + C = 48/15
-------------------------- +
A + 2B = 1/5 + 48/15 = 3/15 + 48/15 = 51/15
4A + 2B = 91/15
A + 2B = 51/15
-------------------------- -
3A = 40
A = 40/3
C = A + 1/6 = 40/3 + 1/6 = 80/6 + 1/6 = 81/6 = 27/2
so the length of rope A is 40/3 m and thr length of rope C is 27/2 m.
10. Twi semicricles are formed by the lenght of a rope as shown.find the length of the rope used.give you answer correct to 1 decimal place
Jawaban:
HAI ADEK.......MAAF BARU TERPINDAI SEKARANG YAH
11. A piece of string is cut into five parts so that the length of each piece forms a number line pattern. if the length of the shortest rope is 10 cm, the rope in the middle is 20 cm and the longest rope is 30 cm, then the original length is...pakai cara ya kak mohon bantuannya :'d
Jawaban:
100 cm
Penjelasan dengan langkah-langkah:
S5 = 5/2(10+30)
S5 = 5/2 × 40
S5 = 100 cm
SEMOGA BERMANFAAT
JADIKAN JAWABAN TERBAIK
12. FIGURE 13.1 shows two boxes P and Q that are being balanced on a uniform board of length 4 m which is pivoted at its centre O. The board has a weight of 100 N while the mass of P and Q are 6 kg and 5 kg, respectively.
Jawaban:
tolong dibantu ya terimakasih
13. Gabriel had 300 cm rope. He cut 7/10 of the rope. What was the length of the rope cut? Pakai cara bentar lagi harus diserahkan tolong kak..
Jawab:
210 cm
Penjelasan dengan langkah-langkah:
Given that
the total length of rope (l) = 300 cm
cut part (n) = 7/10
Find the length of the rope that was cut
= n × l
= 7/10 × 300
= 30×7
= 210 cm
[[ KLF ]]
Jawab:
210 cm
Penjelasan dengan langkah-langkah:
Diketahui : panjang tali = 300 cm
dipotong = 7/10 bagian
Ditanya : panjang yang dipotong = ?
Jika panjang tali = 300 cm
maka 7/10 bagiannya adalah = 300 cm X 7/10
= 210 cm
14. Mrs Lim had a piece of rope 3/4 m long . She cut it into 1/10 m pieces .(a) How many 1 / 10 m long of rope were there at most ? (b) What was the length of the piece of rope left over ?
a) 3/4 ÷ 1/10 = 7 reminder:½
b) ½
15. Rope A is 1/6 meter shorter than Rope B. Rope B is 1/5 meter shorter than Rope C. The total length of the three ropes is 2 8/15 metersA. how long is Rope AB. how long is Rope C
Jawaban:
a + ⅙ = b ... [1]
b + ⅕ = c
substitute B with [1]
a + ⅙ + ⅕ = c
a + 5/30 + 6/30 = c
a + 11/30 = c ... [2]
[tex]a + b + c = 2 \frac{8}{15} [/tex]
substitute b & c with [1] & [2]
[tex]a + a + \frac{1}{6} + a + \frac{11}{30} = 2 \frac{8}{15} [/tex]
[tex]3a + \frac{5}{30} + \frac{11}{30} = \frac{76}{30} [/tex]
[tex]3a = \frac{76 - 5 - 11}{30}[/tex]
[tex]a = \frac{60}{30} \times \frac{1}{3} [/tex]
[tex]a = \frac{2}{3} \: m[/tex]
b.
[tex]c = a + \frac{11}{30} [/tex]
[tex]c = \frac{2}{3} + \frac{11}{30} [/tex]
[tex]c = \frac{20 + 11}{30} [/tex]
[tex]c = \frac{31}{30} [/tex]
[tex]c = 1 \frac{1}{30} \: m[/tex]
16. A cuboid has a volume of 7200cm3,square and a length of 32cm.Gind the length of the square face
Jawaban:
Remember the surface area is the total area of all the faces of a 3D shape.
The lateral surface area of a cuboid is given by:
LSA = 2 (lh + wh) = 2 h (l + w)
Example 1: Find the total surface area of a cuboid with dimensions 8 cm by 6 cm by 5 cm.
TSA = 2 (lw + wh + hl)
TSA = 2 (8*6 + 6*5 + 5*8)
TSA = 2 (48 + 30 + 40)
TSA = 236
So, the total surface area of this cuboid is 236 cm2.
Example 2: Find the surface area of a cuboid of dimensions 4.8 cm, 3.4 cm and 7.2 cm.
Solution:
Area of Face 1: 4.8 × 7.2 = 34.56 cm²
Area of Face 2: 3.4 × 7.2 = 24.48 cm²
Area of Face 3: 4.8 × 3.4 = 16.32 cm²
Adding the area of these 3 faces gives 75.36 cm², since each face is duplicated on the opposite side, the total surface area of the cuboid will be:
TSA = 2 (75.36) = 150.72 cm²
Example 3: The length, width and height of a cuboid are 10cm, 8cm and 7cm respectively. Find the lateral surface area of a cuboid.
Solution:
Lateral surface area of cuboid is given by:
LSA = 2h(l+w)
where,
l = length = 10 cm
w = width = 8 cm
h = height = 7 cm
Insert these values into the formula we will get:
LSA = 2 ×7(10 + 8)
LSA = 14 × 18
LSA = 252 cm2
Example 4: The length, breadth and height of a cuboid are 16cm, 14cm and 10cm respectively. Find the total surface area of the cuboid.
Solution:
The total surface area of a cuboid is given by:
TSA = 2 (l*b + b*h + h*l)
Given that:
l = 16cm
b = 14cm
h = 10cm
Substituting the values in the equation we will get
TSA = 2 (16*4 + 14*10 + 10*16)
TSA = 2(224 + 140 + 160)
TSA = 2 * 524
TSA = 1048 cm2
Example 5: Given a cereal box whose length is 20 cm, height is 30 cm and width is 8 cm. Find the surface area of the box.
Solution:
To find the surface are of the box we need to find the area of each rectangular face and add them all up.
The area of the front face is: 20 x 30 = 600 cm2.
The area of the top face is: 20 x 8 = 160 cm2.
The area of the side face is: 8 x 30 = 240 cm2.
Now add these values together we will get: 600 + 160 + 240 = 1000 cm2.
And the total surface area is therefore 1000 x 2 = 2000 cm2.
Example 6: Find the surface area of a cuboid whose sides are 3cm by 6cm by 10cm.
Solution:
Surface area of the cuboid is given by:
TSA = 2 (16*4 + 14*10 + 10*16)
TSA = 2(3 x 6 + 6 x 10 + 3 x 10)
TSA = 2(18 + 60 + 30)
TSA = 216 cm2
17. 6. Density is a derived quantity are those whose units are derived from bases quantity.... A. length and time B. mass and time C. mass and length D. temperature and timebantuuu yaa..nnti malam terakhir ngumpull
Jawaban:
Density is a derived quantity are those whose units are derived from bases quantity C.massandlength.
18. The original length of a spring is 8 cm. When a mass is attached to it, the length of the spring is 10.5 cm. Find the percentage increase in the length of the spring.
Jawab:
Penjelasan dengan langkah-langkah:
increase in length :
10.5 - 8 = 2.5cm
% increase :
= increase in length / original value x 100%
=2.5 / 8 x 100
=31.25 %
19. A bullet with mass 10 gram is fired and hits a block which mass is 1.49 kg. The block is haning freely on a piece of rope of length with 0.2 m. If the acceleration due to the earth gravity is 10m/s² and the rope is displaced by 60 cm measured from its initial position/vertical position, the velocity of the bullet when fired is ... m/s
Hk Kekekalan Energi
(m+M).g.h = 1/2.(m+M).v'²
v' = √2gh
v' = √(2. 10. 0,6)
v' = 2√3 m/s
Hk Kekekalan Momentum
m.v + M.V = (m+M).v'
0,01. v + 1,49. 0 = (0,01+1,49). 2√3
0,01. v = 1,5. 2√3
v = 300√3 m/s
20. An object has mass of 6kg and 6 m from ground. Calculate the potential energy!
Jawaban:
mass (m) = 6 kg
height (h) = 6 m
gravitational acceleration (g) = 10 m/s²
Potential Energy (Ep) = m×g×h
Ep = 6 kg × 6 m/s² × 10 m
Ep = 360 kg m²/s² = 360 Joule.
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