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OLga [1]
3 years ago
10

Two blocks A and B with mA = 2.6 kg and mB = 0.81 kg are connected by a string of negligible mass. They rest on a frictionless h

orizontal surface. You pull on block A with a horizontal force of 6.5 N. Determine the tension in the string connecting the two blocks.
Physics
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

Explanation:

Given

Mass of block A (m_a)=2.6 kg

Mass of block B (m_b)=0.81 kg

Force of 6.5 N is applied on Block A

Force on Block A is F, Tension

thus

F-T=m_a\times a -----1

where a=acceleration of the system

For Block B

T=m_b\times a -----2

F=m_b\times a+m_a\times a

F=\left ( m_a+m_b\right )a

F=\left ( 2.6+0.81\right )a

a=\frac{6.5}{3.41}

a=1.96 m/s^2

T=0.81\times 1.96=1.54 N

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A 5 cm spring is suspended with a mass of 3.8589 g attached to it which extends the spring by 1.5747 cm. The same spring is plac
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Answer:

charges of the beads is 1.173 ×10^{-15} C

Explanation:

given data

mass = 3.8589 g = 0.003859 kg

spring length = 5 cm = 0.05 m

extend spring x = 1.5747 cm = 0.15747 m

spring's extension = 0.0116 m

to find out

charges of the beads

solution

we know that force is

force = mass × g

force = 0.003859 × 9.8

force = 0.03782 N

so we know  force for mass

force  = -kx

so k = force / x

put here force and x value

k = -0.03782 / 0.1575

k = -0.24 N/m

and

force for spring's extension

force = -kx

force = -0.24 ( 0.0116) = 0.002784 N

so here

total length L = 0.05 + 0.0116 = 0.0616

so charges of the beads = force × L² / ke

charges of the beads = 0.002784 × (0.0616)² / (9 ×10^{9} )

so charges of the beads = 1.173 ×10^{-15} C

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3 years ago
Use the collision theory to explain how increasing the temperature of a reaction will affect the rate of the reaction.
kirill115 [55]
Increasing the temperature causes the particles in the reaction to become kinetically excited, hitting one another in increasing frequency. Increased collision among means faster rate or reaction.
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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

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What force must act on a 50.0-kg mass to give it an ancceleration of 0.30 m/s^2?
ratelena [41]

Answer:

15.0 N

Explanation:

see pic

3 0
2 years ago
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