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2 years ago

Please help! I’ve got 96 N but it’s wrong..

Physics

2 answers:

mash [69]2 years ago

5 0

Dude it’s 8.1 Km. Bet it’s right.

bekas [8.4K]2 years ago

3 0

I’ve got 8.1 km, hope it helps.

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Neil pogo sticks to his science class, but stops to pick up his backpack on his way. He travels 8 m east, then 4 m west. what di

Likurg_2 [28]

Answer:

8.9 m NorthEast

Explanation:

a² + b² = c²

8² + 4² = c²

64 + 16 = c²

80 = c²

c = √80

c ≈ 8.9 m NorthEast

8 0

2 years ago

A 85-kg astronaut is stranded from his space shuttle. He throws a 1-kg hammer away from the shuttle with a velocity of 17 m/s. H

algol [13]

Answer:

0.2 m/s

Explanation:

given,

mass of astronaut, M = 85 Kg

mass of hammer, m = 1 Kg

velocity of hammer , v =17 m/s

speed of astronaut, v' = ?

initial speed of the astronaut and the hammer be equal to zero = ?

Using conservation of momentum

(M + m) V = M v' + m v

(M + m) x 0 = 85 x v' + 1 x 17

85 v' = -17

v' = -0.2 m/s

negative sign represent the astronaut is moving in opposite direction of hammer.

Hence, the speed of the astronaut is equal to 0.2 m/s

4 0

3 years ago

Express newton pascal joule and watt in fundamental unit

Nadya [2.5K]

i added an image. i hope it helped :)

3 0

2 years ago

g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn

krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

T² = ( $\frac{4\pi }{G M_s}$ ) r³ (1)

in this case the period of the season is

T₁ = 93 min (60 s / 1 min) = 5580 s

r₁ = 410 + 6370 = 6780 km

r₁ = 6.780 10⁶ m

for the satellite

T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

T² = K r³

K = T₁²/r₁³

K = $\frac{ 5580^2}{ (6.780 10^6)^2}$

K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

r³ = T² / K

r³ = 86400² / 9.99 10⁻¹⁴

r = ∛(7.4724 10²²)

r = 4.21 10⁷ m

this distance is from the center of the earth

7 0

2 years ago

The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-

nikdorinn [45]

Answer:

The answer is " $60.74^{\circ}$ ".

Explanation:

Cavity and benzene should be extended in equal quantities.

$\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\$

$\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\$

$\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\$

5 0

2 years ago