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8_murik_8 [283]
3 years ago
8

ASAP pls answer right if can’t see picture don’t answer

Physics
1 answer:
Anna35 [415]3 years ago
6 0

Answer:

Not sure but

F = m* a

32= 5 * a

a= 6.4 m/s^2

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In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s
serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

a = \dfrac{1.2 \times 10^5}{86}

a =1395.35\ m/s^2

Using equation of motion

v² = u² + 2 a s

s =\dfrac{v^2}{2a}

s =\dfrac{52^2}{2\times 1395.35}

s = 0.9689 m

The minimum depth of snow that would have stooped him is  s = 0.9689 m

8 0
3 years ago
Use the following equation to help you answer the question. The peak intensity of radiation from a star named Sigma is 2 x 10 6
puteri [66]

Answer:

1449 K

Explanation:

The surface temperature of a star is related to its peak wavelength by Wien's displacement law:

T=\frac{b}{\lambda}

where

T is the surface temperature

b is Wien's displacement constant

\lambda=2\cdot 10^{-6} m

So the surface temperature of the star is

T=\frac{2.898 \cdot 10^{-3} Km}{2\cdot 10^{-6} m}=1449 K

8 0
3 years ago
Read 2 more answers
If a car traveling at 60 km/h will skid 20m when its brakes lock, how far will it skid if it's traveling at 120 km/h when it's b
pantera1 [17]
<span>If a mouse and an elephant both run with the same kinetic energy, can you say ... 3. if a car traveling at60 km/h will skid 20 m when its brakes lock, how far will it skid if it is traveling at 120 km/h when its brakes lock? (This question is typical on ...</span><span>just get this and figure it out and this is your answer </span>
7 0
3 years ago
Please help !! ASAP
umka2103 [35]

Answer:

22) 5.6 Kg.m/s

23) 0.067 Kg

24) 1.16 m/s

Explanation:

Momentum is found by multiplying mass and velocity

P=mv where P is momentum, m is mass of object and v is the velocity

22

Given

Mass, m=2 Kg

Rate, v=2.8 m/s

Relationship

P=mv

Solution

P=mv=2*2.8=5.6 Kg.m/s

23

Given

Momentum, p=0.1 Kg.m/s

Rate, v=1.5 m/s

Relationship

P=mv and making m the subject then

m=\frac {P}{v}

Solution

m=\frac {0.1 Kgm.s}{1.5 m/s}\approx 0.067 Kg

24

Given

Momentum, p=4.9\times 10^{8} Kg.m/s

Mass, m=4.23\times10^{8} Kg

Relationship

P=mv and making v the subject then

v=\frac {P}{m}

Solution

m=\frac {4.9\times 10^{8}  Kgm.s}{4.23\times10^{8} }\approx 1.16 m/s

5 0
2 years ago
Need help answering this question for #30
dezoksy [38]
I think u would add them all up and then write down your answers
6 0
3 years ago
Read 2 more answers
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