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Mashutka [201]
3 years ago
8

Two forces act on a 55-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other ha

s a magnitude 35 N at 32° clockwise from the positive y-axis. What is the magnitude of this objectʹs acceleration? A) 1.1 m/s2 B) 1.3 m/s2 C) 1.5 m/s2
Physics
1 answer:
Gwar [14]3 years ago
6 0

Answer:

A) 1.1 m/s/s

Explanation:

There exist two forces on the object such that

F_1 = 65 N directed 59° clockwise from the positive x-axis

F_2 = 35 N at 32° clockwise from the positive y-axis

now we have

F_1 = 65 cos59\hat i - 65 sin59 \hat j

F_2 = 35 sin32\hat i + 35 cos32 \hat j

now the net force on the object is given as

F_{net} = F_1 + F_2

F_{net} = (65 cos59 + 35 sin32)\hat i + (35cos32 - 65 sin59)\hat j

F_{net} = 52\hat i - 26 \hat j

so it's magnitude is given as

F_{net} = \sqrt{52^2 + 26^2} = 58.15 N

now from Newton's II law we have

F = ma

a = \frac{58.15}{55} = 1.1 m/s^2

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4 years ago
A person is standing on a scale in an unmoving elevator. The elevator starts to move upwards at 1 m/s squared. Is the scale read
timama [110]

Answer: GREATER

Explanation:when elevator does not move it reads weight of the person . when elevator moves up let apparent weight be F . W acts downwards so net force is F-W

HENCE

F-W =ma

F= ma+W

AS a= 1 m/s^2

F = m (1)+W

HENCE GREATER

7 0
3 years ago
8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott
nasty-shy [4]

<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>

<u>Explanation:</u>

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m_1,u_1\text{ and }v_1 are the mass, initial velocity and final velocity of ball.

m_2,u_2\text{ and }v_2 are the mass, initial velocity and final velocity of bottle.

We are given:

m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s

Putting values in above equation, we get:

(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0
3 years ago
Jack travelled 360 km at an average speed of 80 km/h. Elaine
diamong [38]

Answer:

Average speed of Elain = 60 km/h

Explanation:

Total Distance covered by Jack = 360km

Average Speed of Jack = 80 km/h

Time taken by Jack to complete his journey = Distance / Average speed = 360 km / 80 km/h

Time taken by Jack to complete his journey = 4.5 hours

As it is given the both Jack and Elain travelled the same amount of distance:

Total distance travelled by Elain = 360 km

It is given that Elain took 1.5 hourse more than Jack to cover the distance, so Time taken by Elain to cover the distance is = 4.5 hours + 1.5 hours = 6 hours

Average speed of Elain = Distance/ time = 360 km / 6 hours

Average speed of Elain = 60 km/h

3 0
3 years ago
A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density o
IrinaK [193]

Answer:

a. 6.032\times10^{-6}C/m^2

b.6.816\times10^5N/C

Explanation:

#Apply  surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as \sigma

\sigma=\frac{Q}{4\pi r_{out}^2}, and the strength of the electric field outside the sphere E=\frac{\sigma _{new}}{\epsilon _o}

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2  #surface charge outside sphere.

\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2

Hence, the new charge density on the outside of the sphere is 6.032\times10^{-6}C/m^2

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be 6.032\times10^{-6}C/m^2,

E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C

Hence, the strength of the electric field just outside the sphere is 6.816\times10^5N/C

5 0
3 years ago
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