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m_a_m_a [10]
3 years ago
13

I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2

Physics
1 answer:
olga nikolaevna [1]3 years ago
5 0

Answer:

Explanation:

a = \frac{v_f-v_0}{t} which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator.  For us:

a=\frac{2.7-1.6}{1.2} so

a = .92 m/s/s (NOT negative because you're speeding up)

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In a lab, the mass of object a is 52 kg. object a weighs
aksik [14]
If you are asking for the weight then the formula is F=mg where f is weight m is mass and g is acceleration due to gravity.m=52kg and g=9.8m/s2(the gravity of earth)
F=52*9.8=509.6
therefore the weight of the object is 509.6N
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3 years ago
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How is work calculated when the force applied is not parallel to the displacement
Leno4ka [110]
Pretty sure it’s Force*Distance*Cos(theta)
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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
Brrunno [24]

Answer:

The explorer should travel to reach base camp to 5.02 Km at 4.28° south of due west.

Explanation:

Using trigonometric function like Sen(Ф), Cos(Ф) and Tan(Ф) we can get distance and direction that the explorer should travel to reach base camp. When we discompound the vector X = 7.8*Cos(50) = 5.01 Km y y = 7.8 * Sen (50) - 5.6 = 5.975 - 5.6 (Km) = 0.375 (Km) so that Tan (\alpha ) = \frac{0.375}{5.01} = 4.28; \alpha = Arctang (\frac{0.375}{5.01}) = 4.28 (degree) to get how far we use Pythagorean theorem so R^{2} = x^{2}+y^{2} so that R=\sqrt{0.375^{2}+5.01^{2} } =5.02 (Km)

6 0
3 years ago
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When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
A 10 m long uniform beam weighing 100 N is supported by two ropes at the ends. If a 400 N person sits at 2.0 m from one end of t
Naddik [55]

Answer:

T1 = 130N, T2 = 370N

Explanation:

In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.

1. forces:

Let tension in rope 1 be T1 and in rope 2 be T2:

ma = T1 + T2 - 100N - 400N = 0

(1) T1 + T2 = 500N

2. torque around the center point of the beam:

τ = r x F = 5*T1 + 3*400N - 5*T2 = 0

(2) T1 - T2 = -240N

Solving both equations:

T1 = 130N

T2 = 370N

3 0
4 years ago
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